Solutions
Consider the following aqueous solutions.
I. 2.2 g Glucose in 125 mL of solution.
II. 1.9 g Calcium chloride in 250 mL of solution.
III. 9.0 g Urea in 500 mL of solution.
IV. 20.5 g Aluminium sulphate in 750 mL of solution.
The correct increasing order of boiling point of these solutions will be :
[Given : Molar mass in g mol−1 : H = 1, C = 12, N = 14, O = 16, Cl = 35.5, Ca = 40, Al = 27 and S = 32]
III < I < II < IV
II < III < IV < I
II < III < I < IV
I < II < III < IV
Explanation:
$ \begin{aligned} &\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{i} \cdot \mathrm{k}_{\mathrm{b}} \cdot \mathrm{~m}\\ &\text { For dilute solution }(M=m) \end{aligned} $
$\text{We know that boiling point elevation depends on } \Delta \mathrm{T}_{\mathrm{b}}.$
$\text{Since } \mathrm{k}_\mathrm{b} \text{ is same for water in all cases, we compare only } \mathrm{i}\times m \text{ (or } \mathrm{i}\times M\text{ for dilute solutions).}$
| $ \text { Molarity } $ |
$ \mathbf{i} \times \mathbf{m} $ |
|---|---|
| $ \text { (I) } \mathrm{M}_{\text {glucose }}=\frac{2.2}{180} \times \frac{1000}{125}=0.098 $ |
$ 0.098 \times 1 $ |
| $ \text { (II) } \mathrm{M}_{\mathrm{CaCl}_2}=\frac{1.9}{111} \times \frac{1000}{250}=0.068 $ |
$ 0.068 \times 3 $ |
| $ \text { (III) } \mathrm{M}_{\text {urea }}=\frac{9}{60} \times \frac{1000}{500}=0.3 $ |
$ 0.3 \times 1 $ |
| $ \text { (IV) } \mathrm{M}_{\mathrm{Al}_2\left(\mathrm{SO}_4\right)_2}=\frac{20.5}{342} \times \frac{1000}{750} \simeq 0.08 $ |
$ 0.08 \times 5 $ |
$\text{Now compare the values of } \mathrm{i}\times m:$
$\text{(I) Glucose: } 0.098\times 1=0.098$
$\text{(II) } \mathrm{CaCl}_2\text{: } 0.068\times 3=0.204$
$\text{(III) Urea: } 0.3\times 1=0.3$
$\text{(IV) Aluminium sulphate: } 0.08\times 5=0.4$
Order of $\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{Al}_2\left(\mathrm{SO}_4\right)_3>$ Urea $>\mathrm{CaCl}_2>$ Glucose
So order of BP $=\mathrm{Al}_2\left(\mathrm{SO}_4\right)_3>$ Urea $>\mathrm{CaCl}_2>$ Glucose
So Answer will be I $<$ II $<$ III $<$ IV
At $\mathrm{T}(\mathrm{K}), 2$ moles of liquid A and 3 moles of liquid B are mixed. The vapour pressure of ideal solution formed is 320 mm Hg . At this stage, one mole of A and one mole of B are added to the solution. The vapour pressure is now measured as 328.6 mm Hg . The vapour pressure (in mm Hg ) of A and B are respectively:
400, 300
600, 400
500, 200
300, 200
Explanation:
For an ideal solution, by Raoult’s law:
$P_{\text{total}}=x_A P_A^0 + x_B P_B^0$
where $P_A^0,\;P_B^0$ are vapour pressures of pure liquids A and B.
Step 1: First mixture (2 mol A + 3 mol B)
Total moles $=2+3=5$
$x_A=\frac{2}{5},\quad x_B=\frac{3}{5}$
Given $P_{\text{total}}=320$ mm Hg:
$320=\frac{2}{5}P_A^0+\frac{3}{5}P_B^0$
Multiply by 5:
$1600=2P_A^0+3P_B^0 \quad ...(1)$
Step 2: After adding 1 mol A and 1 mol B
New moles: A $=3$, B $=4$, total $=7$
$x_A=\frac{3}{7},\quad x_B=\frac{4}{7}$
Given new vapour pressure $=328.6$ mm Hg:
$328.6=\frac{3}{7}P_A^0+\frac{4}{7}P_B^0$
Multiply by 7:
$2300.2=3P_A^0+4P_B^0 \quad ...(2)$
Step 3: Solve the two equations
From (1): $2P_A^0+3P_B^0=1600$
From (2): $3P_A^0+4P_B^0=2300.2$
Multiply (1) by 3:
$6P_A^0+9P_B^0=4800$
Multiply (2) by 2:
$6P_A^0+8P_B^0=4600.4$
Subtract:
$P_B^0=4800-4600.4=199.6 \approx 200$
Put in (1):
$2P_A^0+3(199.6)=1600$
$2P_A^0+598.8=1600$
$2P_A^0=1001.2 \Rightarrow P_A^0=500.6 \approx 500$
Final Answer
Vapour pressures of A and B are approximately:
$P_A^0=500\ \text{mm Hg},\quad P_B^0=200\ \text{mm Hg}$
Correct option: C (500, 200)
At 298 K , the mole percentage of $\mathrm{N}_2(\mathrm{~g})$ in air is $80 \%$. Water is in equilibrium with air at a pressure of 10 atm . What is the mole fraction of $\mathrm{N}_2(\mathrm{~g})$ in water at 298 K ? $\left(\mathrm{K}_{\mathrm{H}}\right.$ for $\mathrm{N}_2$ is $\left.6.5 \times 10^7 \mathrm{~mm} \mathrm{Hg}\right)$
$9.35 \times 10^{-5}$
$9.35 \times 10^5$
$1.23 \times 10^{-7}$
$1.17 \times 10^{-4}$
Explanation:
$ \begin{aligned} & \mathrm{P}_{\mathrm{N}_2}=\mathrm{K}_{\mathrm{H}} \cdot \mathrm{X}_{\mathrm{N}_2} \\ & \mathrm{P}_{\mathrm{N}_2}=0.8 \times 10=8 \mathrm{~atm} \\ & 8 \times 760=6.5 \times 10^7 \times \mathrm{X}_{\mathrm{N}_2} \\ & \mathrm{X}_{\mathrm{N}_2}=\frac{8 \times 760}{6.5 \times 10^7} \\ & \mathrm{X}_{\mathrm{N}_2}=9.35 \times 10^{-5} \end{aligned} $
Two liquids A and B form an ideal solution at temperature T K . At T K , the vapour pressures of pure A and B are 55 and $15 \mathrm{kN} \mathrm{m}^{-2}$ respectively. What is the mole fraction of A in solution of A and B in equilibrium with a vapour in which the mole fraction of A is 0.8?
0.5217
0.480
0.340
0.663
Explanation:
For an ideal solution, Raoult’s law applies:
$ p_A=x_A P_A^*,\qquad p_B=x_B P_B^* $
Total pressure:
$ P=p_A+p_B=55x_A+15(1-x_A)=55x_A+15-15x_A=40x_A+15 $
Vapour-phase mole fraction (Dalton’s law):
$ y_A=\frac{p_A}{P}=\frac{55x_A}{40x_A+15}=0.8 $
Solve:
$ 55x_A=0.8(40x_A+15)=32x_A+12 $
$ 23x_A=12 \Rightarrow x_A=\frac{12}{23}=0.5217 $
Answer: 0.5217 (Option A)
A solution is prepared by dissolving 0.3 g of a non-volatile non-electrolyte solute 'A' of molar mass $60 \mathrm{~g} \mathrm{~mol}^{-1}$ and 0.9 g of a non-volatile non-electrolyte solute ' B ' of molar mass $180 \mathrm{~g} \mathrm{~mol}^{-1}$ in $100 \mathrm{~mL} \mathrm{H}_2 \mathrm{O}$ at $27^{\circ} \mathrm{C}$. Osmotic pressure of the solution will be
[Given: $\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ ]
2.46 atm
0.82 atm
1.47 atm
1.23 atm
Explanation:
Moles of solute $A$:
$n_A=\frac{0.3}{60}=0.005\ \text{mol}$
Moles of solute $B$:
$n_B=\frac{0.9}{180}=0.005\ \text{mol}$
Total moles of solute particles (both are non-electrolytes, so no dissociation):
$n=n_A+n_B=0.005+0.005=0.01\ \text{mol}$
Volume of solution (dilute solution, take volume $\approx 100\ \text{mL}=0.1\ \text{L}$):
$V=0.1\ \text{L}$
Temperature:
$T=27^\circ\text{C}=300\ \text{K}$
Osmotic pressure:
$\pi=\frac{n}{V}RT=\frac{0.01}{0.1}\times 0.082 \times 300$
$\pi=0.1\times 24.6=2.46\ \text{atm}$
Correct option: A) $2.46\ \text{atm}$
' W ' g of a non-volatile electrolyte solid solute of molar mass ' M ' $\mathrm{g} \mathrm{mol}^{-1}$ when dissolved in 100 mL water, decreases vapour pressure of water from 640 mm Hg to 600 mm Hg . If aqueous solution of the electrolyte boils at 375 K and $\mathrm{K}_{\mathrm{b}}$ for water is $0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$, then the mole fraction of the electrolyte solute $\left(x_2\right)$ in the solution can be expressed as
(Given : density of water $=1 \mathrm{~g} / \mathrm{mL}$ and boiling point of water $=373 \mathrm{~K}$ )
$\frac{16}{2 \cdot 6} \times \frac{W}{M}$
$\frac{1 \cdot 3}{8} \times \frac{\mathrm{M}}{\mathrm{W}}$
$\frac{2 \cdot 6}{16} \times \frac{\mathrm{M}}{\mathrm{W}}$
$\frac{1 \cdot 3}{8} \times \frac{W}{M}$
Explanation:
Boiling point elevation:
$ \Delta T_b = 375-373 = 2\,\text{K} $
Mass of water $=100\,\text{mL}\times 1\,\text{g mL}^{-1}=100\,\text{g}=0.1\,\text{kg}$
Molality:
$ m=\frac{\text{moles of solute}}{\text{kg of solvent}} =\frac{(W/M)}{0.1}=10\frac{W}{M} $
For electrolyte (NCERT):
$ \Delta T_b = iK_b m $
$ 2=i(0.52)\left(10\frac{W}{M}\right)=i\left(5.2\frac{W}{M}\right) $
$ i=\frac{2}{5.2}\cdot\frac{M}{W}=\frac{5}{13}\cdot\frac{M}{W} $
Vapour pressure lowering:
$ \frac{\Delta p}{p^\circ}=\frac{640-600}{640}=\frac{40}{640}=\frac{1}{16} $
For electrolyte (NCERT):
$ \frac{\Delta p}{p^\circ}=ix_2 $
So,
$ x_2=\frac{1/16}{i}=\frac{1}{16}\cdot\frac{1}{\left(\frac{5}{13}\cdot\frac{M}{W}\right)} =\frac{1}{16}\cdot\frac{13}{5}\cdot\frac{W}{M} =\frac{13}{80}\cdot\frac{W}{M} $
$ x_2=\left(\frac{1.3}{8}\right)\frac{W}{M} $
Correct option: D
Which one of the following graphs accurately represents the plot of partial pressure of $\mathrm{CS}_2$ vs its mole fraction in a mixture of acetone and $\mathrm{CS}_2$ at constant temperature?
Explanation:
Mixture of $\mathrm{CS}_2$ and
show positive deviation $\mathrm{P}_{\mathrm{CS}_2}>\mathrm{P}_{\mathrm{CS}_2}^{\mathrm{o}} \cdot \mathrm{X}_{\mathrm{CS}_2}$
At $\mathrm{T}(\mathrm{K}), 100 \mathrm{~g}$ of $98 \% \mathrm{H}_2 \mathrm{SO}_4(\mathrm{w} / \mathrm{w})$ aqueous solution is mixed with 100 g of $49 \% \mathrm{H}_2 \mathrm{SO}_4(\mathrm{w} / \mathrm{w})$ aqueous solution. What is the mole fraction of $\mathrm{H}_2 \mathrm{SO}_4$ in the resultant solution?
(Given : Atomic mass $\mathrm{H}=1 \mathrm{u} ; \mathrm{S}=32 \mathrm{u} ; \mathrm{O}=16 \mathrm{u}$ ).
(Assume that temperature after mixing remains constant)
0.9
0.663
0.1
0.337
Explanation:
Total mass of $\mathrm{H}_2\mathrm{SO}_4$ in the mixture:
$ =\left(100 \times \frac{98}{100}\right)+\left(100 \times \frac{49}{100}\right)=147 \mathrm{gm} $
Here, the first solution gives $98\ \mathrm{g}$ of $\mathrm{H}_2\mathrm{SO}_4$ and the second gives $49\ \mathrm{g}$ of $\mathrm{H}_2\mathrm{SO}_4$, so total is $147\ \mathrm{g}$.
Total mass of solution after mixing $=100+100=200\ \mathrm{g}$.
So, total mass of $\mathrm{H}_2\mathrm{O}$ in the mixture:
Total weight of $\mathrm{H}_2 \mathrm{O}=200-147=53 \mathrm{gm}$
Moles of $\mathrm{H}_2\mathrm{SO}_4 = \frac{147}{98}$ (since molar mass of $\mathrm{H}_2\mathrm{SO}_4$ is $98\ \mathrm{g\,mol^{-1}}$).
Moles of $\mathrm{H}_2\mathrm{O} = \frac{53}{18}$ (since molar mass of $\mathrm{H}_2\mathrm{O}$ is $18\ \mathrm{g\,mol^{-1}}$).
Now, mole fraction of $\mathrm{H}_2 \mathrm{SO}_4$ is:
Mole fraction of $\mathrm{H}_2 \mathrm{SO}_4=\frac{\frac{147}{98}}{\left(\frac{147}{98}+\frac{53}{18}\right)}=0.337$
Consider a solution of $\mathrm{CO}_2(\mathrm{~g})$ dissolved in water in a closed container.
Which one of the following plots correctly represents variation of log (partial pressure of $\mathrm{CO}_2$ in vapour phase above water) $[y$-axis $]$ with $\log$ (mole fraction of $\mathrm{CO}_2$ in water) $[x$-axis $]$ at $25^{\circ} \mathrm{C}$ ?
Explanation:
From Henry’s law (as given in NCERT), the partial pressure of a gas above the solution is directly proportional to its mole fraction in the solution.
So, we write:
$ \begin{aligned} & \mathrm{P}(\mathrm{~g})=\mathrm{K}_{\mathrm{H}} \cdot \mathrm{X}(\mathrm{~g}) \\\\ & \operatorname{logP}(\mathrm{g})=\log \mathrm{K}_{\mathrm{H}}+\operatorname{logX}(\mathrm{g}) \end{aligned} $
Now compare this with the straight-line form $y = c + x$.
Here, $y = \log P(\mathrm{g})$, $x = \log X(\mathrm{g})$, and the constant (intercept) is $\log K_H$.
Therefore, the graph between $\log P$ (on $y$-axis) and $\log X$ (on $x$-axis) will be a straight line with slope $1$ and $y$-intercept $\log K_H$.
Given below are two statements :
Statement I : The Henry's law constant $\mathrm{K}_{\mathrm{H}}$ is constant with respect to variations in solution's concentration over the range for which the solution is ideally dilute.
Statement II : $\mathrm{K}_{\mathrm{H}}$ does not differ for the same solute in different solvents.
In the light of the above statements, choose the correct answer from the options given below
Both Statement I and Statement II are true
Statement I is true but Statement II is false
Statement I is false but Statement II is true
Both Statement I and Statement II are false
Explanation:
Statement I: True.
For an ideally dilute solution, Henry’s law is obeyed:
$p = K_H x$
At a fixed temperature, $K_H$ remains constant over the concentration range where the solution behaves ideally (i.e., dilute enough).
Statement II: False.
$K_H$ depends on the nature of the gas (solute) and also the nature of the solvent (and temperature). So, for the same solute, $K_H$ can be different in different solvents.
Correct option: B (Statement I is true but Statement II is false).
Elements P and Q form two types of non-volatile, non-ionizable compounds PQ and $\mathrm{PQ}_2$. When 1 g of $P Q$ is dissolved in 50 g of solvent ' $A^{\prime}, \Delta T_b$ was 1.176 K while when 1 g of $P Q_2$ is dissolved in 50 g of solvent ' $\mathrm{A}^{\prime}, \Delta \mathrm{T}_{\mathrm{b}}$ was 0.689 K . ( $\mathrm{K}_{\mathrm{b}}$ of ' $\mathrm{A}^{\prime}=5 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ ). The molar masses of elements P and Q (in $\mathrm{g} \mathrm{mol}^{-1}$ ) respectively, are :
25, 60
60, 25
65, 145
70, 110
Explanation:
For a non-volatile, non-ionizable solute,
$\Delta T_b = K_b \, m$
and
$m=\frac{\text{moles of solute}}{\text{kg of solvent}}$
Given: mass of solvent $=50\,\text{g}=0.05\,\text{kg}$, mass of solute $=1\,\text{g}$.
So for any solute of molar mass $M$,
$ m=\frac{\frac{1}{M}}{0.05}=\frac{1}{0.05M}=\frac{20}{M} $
Hence,
$ \Delta T_b=K_b \cdot \frac{20}{M}=5\cdot \frac{20}{M}=\frac{100}{M} $
So,
$ M=\frac{100}{\Delta T_b} $
1) For $PQ$
$ M(PQ)=\frac{100}{1.176}\approx 85.03 \approx 85 $
Let molar masses of elements be $P=x$ and $Q=y$:
$ x+y=85 \quad ...(1) $
2) For $PQ_2$
$ M(PQ_2)=\frac{100}{0.689}\approx 145.14 \approx 145 $
So,
$ x+2y=145 \quad ...(2) $
Subtract (1) from (2):
$ (x+2y)-(x+y)=145-85 \Rightarrow y=60 $
Then from (1):
$ x+60=85 \Rightarrow x=25 $
Therefore, molar masses are:
$ P=25,\quad Q=60 $
Correct option: A (25, 60)
Two liquids A and B form an ideal solution. At 320 K , the vapour pressure of the solution, containing 3 mol of $A$ and 1 mol of $B$ is 500 mm Hg . At the same temperature, if 1 mol of A is further added to this solution, vapour pressure of the solution increases by 20 mm Hg . Vapour pressure (in mm Hg ) of B in pure state is $\_\_\_\_$ . (Nearest integer)
Explanation:
For an ideal solution, by Raoult’s law:
$P_{\text{total}}=x_A P_A^0 + x_B P_B^0$
First solution (3 mol $A$, 1 mol $B$)
Total moles $=3+1=4$
$x_A=\frac{3}{4},\quad x_B=\frac{1}{4}$
Given $P_{\text{total}}=500$ mm Hg:
$500=\frac{3}{4}P_A^0+\frac{1}{4}P_B^0$
Multiply by 4:
$2000=3P_A^0+P_B^0 \quad ...(1)$
After adding 1 mol $A$ (4 mol $A$, 1 mol $B$)
Total moles $=4+1=5$
$x_A=\frac{4}{5},\quad x_B=\frac{1}{5}$
New vapour pressure $=500+20=520$ mm Hg:
$520=\frac{4}{5}P_A^0+\frac{1}{5}P_B^0$
Multiply by 5:
$2600=4P_A^0+P_B^0 \quad ...(2)$
Solve
Subtract (1) from (2):
$2600-2000=(4P_A^0+P_B^0)-(3P_A^0+P_B^0)$
$600=P_A^0$
Put in (1):
$2000=3(600)+P_B^0=1800+P_B^0$
$P_B^0=200 \text{ mm Hg}$
Answer: $200$
The osmotic pressure of a living cell is 12 atm at 300 K. The strength of sodium chloride solution that is isotonic with the living cell at this temperature is ________ g L-1. (Nearest integer)
Given : R = 0.08 L atm K-1 mol-1
Assume complete dissociation of NaCl
(Given: Molar mass of Na and Cl are 23 and 35.5 g mol-1 respectively.)
Explanation:
Osmotic pressure ($\pi$) = $12 \text{ atm}$
Temperature ($T$) = $300 \text{ K}$
Gas constant ($R$) = $0.08 \text{ L atm K}^{-1} \text{ mol}^{-1}$
Sodium chloride (NaCl) is an electrolyte that breaks into ions in an aqueous solution:
$ \text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^- $
Since it is given to assume complete dissociation, one unit of NaCl produces 2 ions.
Therefore, the van 't Hoff factor, $i = 2$.
According to the formula for osmotic pressure:
$ \pi = iCRT $
Substitute the given values into the equation:
$ 12 = 2 \times C \times 0.08 \times 300 $
$ 12 = 2 \times 24 \times C $
$ 12 = 48 \times C $
$ C = \frac{12}{48} $
$ C = 0.25 \text{ mol L}^{-1} $
Molar mass of NaCl = Molar mass of Na + Molar mass of Cl
$ \text{Molar mass of NaCl} = 23 + 35.5 = 58.5 \text{ g mol}^{-1} $
To convert the concentration from moles per liter to grams per liter (strength), multiply the molar concentration ($C$) by the molar mass:
$ \text{Strength} (\text{g L}^{-1}) = \text{Molar concentration} \times \text{Molar mass} $
$ \text{Strength} = 0.25 \text{ mol L}^{-1} \times 58.5 \text{ g mol}^{-1} $
$ \text{Strength} = 14.625 \text{ g L}^{-1} $
Rounding off to the nearest integer, we get $15$.
The strength of the sodium chloride solution is 15 $\text{g L}^{-1}$.
A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol−1) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is ______ × 10−2. (nearest integer)
[Given: Kb of the solvent = 5.0 K kg mol−1]
Assume the solution to be dilute and no association or dissociation of X takes place in solution.
Explanation:
$ \begin{aligned} & \Delta \mathrm{T}_{\mathrm{b}}=\mathrm{i} \times \mathrm{K}_{\mathrm{b}} \times \mathrm{m} \\ & 0.5=\mathrm{i} \times \mathrm{m} \times 5 \\ & \mathrm{i} \times \mathrm{m}=\frac{0.5}{5}=0.1 \\ & \mathrm{i} \times \mathrm{a}=\frac{15}{1000} \end{aligned} $
(where $\mathrm{a}=$ moles of solute)
Now,
$\begin{aligned} & \frac{P_o-P_S}{P^o}=i X_{\text {solute }}=i \times \frac{a}{a+\frac{150}{300}} \\ & =i \times \frac{a}{1 / 2}=\frac{15 / 1000}{1 / 2}=\frac{30}{1000}=3 \times 10^{-2}=3\end{aligned}$
In a solvent $\mathbf{S}$, a compound $\mathbf{B}$ is partially dissociated into $\mathbf{C}$ and $\mathbf{D}$ as given below :
$ \mathbf{B} \rightleftharpoons 2 \mathbf{C}+2 \mathbf{D} $
$\mathbf{B}, \mathbf{C}$ and $\mathbf{D}$ are non-volatile in nature. The molar mass of $\mathbf{B}$ is 10 times the molar mass of $\mathbf{S}$. The standard boiling point and the standard enthalpy of vaporization of $\mathbf{S}$ are 400 K and $10 R \mathrm{~J} \mathrm{~mol}^{-1}$, respectively ( $R$ is the gas constant in $\mathrm{J} \mathrm{K}^{-1} \mathrm{~mol}^{-1}$ ). A solution of $\mathbf{B}$ in $\mathbf{S}$ with an initial concentration of $\mathbf{B}$ as $0.25 \%$ (mass/mass) has a boiling point of 408 K at 1 bar pressure. In this solution, the mole percent of $\mathbf{B}$ that has been dissociated is $\_\_\_\_$ .
Explanation:
The dissociation of the compound is given as:
$\mathrm{B} \rightleftharpoons 2\mathrm{C} + 2\mathrm{D}$
Here, one molecule of B produces a total of 4 particles (2 of C and 2 of D). Hence,
$n = 4$
The van’t Hoff factor for partial dissociation is given by:
$i = 1 + (n - 1)\alpha$
Substituting $ n = 4 $:
$i = 1 + 3\alpha$
Given that the solution contains 0.25% (mass/mass) of B, it means that in 100 g of solution, there are 0.25 g of B and 99.75 g of solvent S.
Molality $m$ is given by:
$m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} = \frac{\frac{0.25}{M_{\mathrm{B}}}}{0.09975} = \frac{1000}{399\,M_{\mathrm{B}}}$
The elevation in boiling point is:
$\Delta T_b = 408 - 400 = 8\,K$
The formula for the boiling point elevation constant $K_b$ is:
$K_b = \frac{R(T_b^{\circ})^2 M_{\mathrm{S}}}{1000\,\Delta H_{\mathrm{vap}}}$
Substitute given values: $ T_b^{\circ} = 400\,K $, $ \Delta H_{\mathrm{vap}} = 10R $
$K_b = \frac{R \times (400)^2 \times M_{\mathrm{S}}}{1000 \times 10R}$
$K_b = 16\,M_{\mathrm{S}}$
Now, the relation between the elevation in boiling point and molality is:
$\Delta T_b = i \times K_b \times m$
Substitute the known values:
$8 = (1 + 3\alpha) \times 16\,M_{\mathrm{S}} \times \frac{1000}{399\,M_{\mathrm{B}}}$
Given that $ M_{\mathrm{B}} = 10\,M_{\mathrm{S}} $:
$8 = (1 + 3\alpha) \times 16\,M_{\mathrm{S}} \times \frac{1000}{399 \times 10\,M_{\mathrm{S}}}$
On simplifying:
$1.995 = 1 + 3\alpha$
$\alpha = 0.3316$
Therefore, the percentage dissociation is:
$\%\,\alpha = 33.16\%$
Two volatile liquids $\mathbf{A}$ and $\mathbf{B}$ form an ideal solution. Consider a 5 molal solution of $\mathbf{B}$ in $\mathbf{A}$ inside a closed container having a total vapour pressure of 100 mm Hg at 300 K . The vapour pressure of pure $\mathbf{A}$ at 300 K is 105 mm Hg . Assume that $\mathbf{A}$ and $\mathbf{B}$ behave as ideal gases in the vapour phase.
Given :
The gas constant $R=0.08 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$
Molar mass of $\mathbf{A}$ is $50 \mathrm{~g} \mathrm{~mol}^{-1}$
Molar mass of $\mathbf{B}$ is $57 \mathrm{~g} \mathrm{~mol}^{-1}$
Density of liquid $\mathbf{B}$ at 300 K is $0.5 \mathrm{~g} / \mathrm{mL}$
$1 \mathrm{~atm}=760 \mathrm{~mm} \mathrm{Hg}$
At 300 K , the ratio of the molar volume of pure $\mathbf{B}$ in vapour phase to its molar volume in liquid phase is $\_\_\_\_$ .
Explanation:
For an ideal solution, Raoult’s Law gives the total vapour pressure as:
$ P_T = X_A P_A^o + X_B P_B^o $
We are told that the total vapour pressure is 100 mm Hg, and the vapour pressure of pure A is 105 mm Hg.
Since the solution is 5 molal (that is, 5 moles of B dissolved in 1 kg of A), the number of moles of A is:
$ n_A = \frac{1000}{50} = 20 \text{ mol} $
Now, the mole fractions can be written as:
$ X_A = \frac{20}{20 + 5}, \quad X_B = \frac{5}{20 + 5} $
Putting these values into Raoult’s Law:
$ 100 = \frac{20}{25}(105) + \frac{5}{25}(P_B^o) $
Solving for $ P_B^o $:
$ 100 = 84 + 0.2 P_B^o $
$ P_B^o = 80 \text{ mm Hg} $
Molar volume of B in vapour phase
Using the ideal gas law, the molar volume is:
$ V_m = \frac{RT}{P} $
Substituting the given values:
$ V_m = \frac{0.08 \times 300}{80/760} $
$ V_m = 228 \text{ L mol}^{-1} $
Molar volume of B in liquid phase
Molar volume = $ \frac{\text{Molar mass}}{\text{Density}} $
$ V_m' = \frac{57}{0.5} = 114 \text{ mL mol}^{-1} $
Ratio of molar volumes
$ \text{Ratio} = \frac{\text{Molar volume in vapour phase}}{\text{Molar volume in liquid phase}} $
$ \text{Ratio} = \frac{228 \times 1000}{114} = 2000 $
Two volatile liquids $\mathbf{A}$ and $\mathbf{B}$ form an ideal solution. Consider a 5 molal solution of $\mathbf{B}$ in $\mathbf{A}$ inside a closed container having a total vapour pressure of 100 mm Hg at 300 K . The vapour pressure of pure $\mathbf{A}$ at 300 K is 105 mm Hg . Assume that $\mathbf{A}$ and $\mathbf{B}$ behave as ideal gases in the vapour phase.
Given :
The gas constant $R=0.08 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$
Molar mass of $\mathbf{A}$ is $50 \mathrm{~g} \mathrm{~mol}^{-1}$
Molar mass of $\mathbf{B}$ is $57 \mathrm{~g} \mathrm{~mol}^{-1}$
Density of liquid $\mathbf{B}$ at 300 K is $0.5 \mathrm{~g} / \mathrm{mL}$
$1 \mathrm{~atm}=760 \mathrm{~mm} \mathrm{Hg}$
The mole fraction of $\mathbf{B}$ in vapour phase which is in equilibrium with this solution is $\_\_\_\_$ .
Explanation:
For an ideal solution, the total vapour pressure is given by Raoult’s law:
$ P_T = X_A P_A^o + X_B P_B^o $
Given total vapour pressure, $P_T = 100\ \mathrm{mm\ Hg}$, and vapour pressure of pure A, $P_A^o = 105\ \mathrm{mm\ Hg}$.
We have a 5 molal solution, meaning 5 moles of solute B are mixed with 1 kg (1000 g) of solvent A. Moles of A are:
$ n_A = \frac{1000}{50} = 20 $
Total moles in solution = $ n_A + n_B = 20 + 5 = 25 $.
Now applying Raoult’s law:
$ 100 = \frac{20}{25}(105) + \frac{5}{25}(P_B^o) $
Solving for $ P_B^o $:
$ P_B^o = 80\ \mathrm{mm\ Hg} $
Molar volume of vapour B:
For an ideal gas, $ V_m = \frac{RT}{P} $.
$ V_m = \frac{0.08 \times 300}{80 / 760} $
$ V_m = 228\ \text{L/mol} $
Partial pressure of B in the mixture:
According to Raoult’s law,
$ P_B = x_B P_B^o $
Since $ x_B = \frac{5}{25} = 0.2 $,
$ P_B = 0.2 \times 80 = 16\ \mathrm{mm\ Hg} $
The vapour-phase mole fraction of B is:
$ y_B = \frac{P_B}{P_T} = \frac{16}{100} = 0.16 $
Molar volume of liquid B:
$ V_m' = \frac{57}{0.5} = 114\ \mathrm{mL/mol} $
Ratio of molar volumes:
$ \text{Ratio} = \frac{\text{molar volume in vapour}}{\text{molar volume in liquid}} = \frac{228 \times 1000}{114} = 2000 $
Hence, the mole fraction of B in the vapour phase is:
$ y_B = 0.16 $
Given below are two statements :
Given : Molar mass of $\mathrm{C}, \mathrm{H}, \mathrm{O}, \mathrm{Cl}$ are $12,1,16$ and $35.5 \mathrm{~g} \mathrm{~mol}^{-1}$, respectively
Statement I : In $30 \%(\mathrm{w} / \mathrm{w})$ solution of methanol in $\mathrm{CCl}_4($ at T K$)$, the mole fraction of $\mathrm{CCl}_4$ is equal to 0.33 .
Statement II : Mixture of methanol and $\mathrm{CCl}_4$ shows positive deviation from Raoult's law. In the light of the above statements, choose the correct answer from the options given below :
Both Statement I and Statement II are true
Both Statement I and Statement II are false
Statement I is true but Statement II is false
Statement I is false but Statement II is true
Explanation:
Molar masses:
C = 12 g mol⁻¹
H = 1 g mol⁻¹
O = 16 g mol⁻¹
Cl = 35.5 g mol⁻¹
Formula masses:
For methanol, CH₃OH:
$ M = 12 + (3 \times 1) + 16 + 1 = 32\ \text{g mol}^{-1} $
For carbon tetrachloride, CCl₄:
$ M = 12 + (4 \times 35.5) = 12 + 142 = 154\ \text{g mol}^{-1} $
This means in 100 g of solution:
Mass of methanol = 30 g
Mass of CCl₄ = 70 g
$ n_{\text{methanol}} = \frac{30}{32} = 0.9375\ \text{mol} $
$ n_{\text{CCl}_4} = \frac{70}{154} = 0.4545\ \text{mol} $
Total moles = $ 0.9375 + 0.4545 = 1.392 $
So,
$ x_{\text{methanol}} = \frac{0.9375}{1.392} = 0.674 $
$ x_{\text{CCl}_4} = \frac{0.4545}{1.392} = 0.326 \approx 0.33 $
Hence, Statement I is correct.
Methanol is polar and forms hydrogen bonds.
Carbon tetrachloride is non-polar.
When they are mixed, the H-bonding between methanol molecules weakens (since CCl₄ cannot form hydrogen bonds with methanol).
→ The intermolecular attraction between unlike molecules decreases.
As a result, the escaping tendency of molecules increases and vapour pressure increases, leading to positive deviation from Raoult’s law.
Hence, Statement II is also correct.
Final Answer
Both Statement I and Statement II are true.
$ \boxed{\text{Option A}} $
Which of the following statements are not correct?
A. For water, magnitude of $\mathrm{K}_{\mathrm{b}}$ is more than the magnitude of $\mathrm{K}_{\mathrm{f}}$.
B. The elevation in boiling point of water when a non-volatile solute is added to it is larger in magnitude than its depression in freezing point.
C. Osmotic pressure measurement is preferred over any other colligative property to determine molar mass of proteins and polymers.
$ \text { D. The dimerised form of benzoic acid in benzene is } $ 
Choose the correct answer from the options given below :
A and B only
A and D only
A, B and D only
A, C and D only
Explanation:
(A) $\mathrm{K}_{\mathrm{b}}$ for water $=0.512 \frac{{ }^{\circ} \mathrm{C}-\mathrm{Kg}}{\mathrm{mol}}$
$ \mathrm{K}_{\mathrm{f}} \text { for water }=1.86 \frac{{ }^{\circ} \mathrm{C}-\mathrm{Kg}}{\mathrm{~mol}} $
So, $\mathrm{K}_{\mathrm{b}}<\mathrm{K}_{\mathrm{f}}$ for water
(B) Since, $\mathrm{K}_{\mathrm{b}}<\mathrm{K}_{\mathrm{f}}$ so $\Delta \mathrm{T}_{\mathrm{b}}<\Delta \mathrm{T}_{\mathrm{f}}$ [same equimolal solution]
(C) Osmotic pressure is used to determine molar mass of protiens \& polymers
(D) dimerised form of benzoic acid in benzene
Option (A), (B) & (D) are incorrect
Given below are two statements :
Statement I : $\quad \mathrm{H}_2 \mathrm{O}$ molecules move from the chamber 1 to chamber 2 .
Statement II : The osmotic pressure of a solution prepared by dissolving 50 mg of potassium sulphate (molar mass $=174 \mathrm{~g} / \mathrm{mol}$ ) in 2 L of water (at $27^{\circ} \mathrm{C}$ ) is 0.0107 bar. (Given: $\mathrm{R}=0.083 \mathrm{dm}^3$ bar $\mathrm{K}^{-1} \mathrm{~mol}^{-1}$ and assume complete dissociation of electrolyte)
In the light of the above statements, choose the correct answer from the options given below :
Both Statement I and Statement II are true
Both Statement I and Statement II are false
Statement I is true but Statement II is false
Statement I is false but Statement II is true
Explanation:
In chamber (I), the concentration of solute is calculated as:
$C_1 = \frac{18/180}{0.1} = 1\, \text{M}$
In chamber (II), the concentration is given by:
$C_2 = \frac{30/180}{250/1000} = \frac{2}{3}\, \text{M}$
Since osmotic pressure depends on concentration and $\pi_1 > \pi_2$, water will move from the side having lower osmotic pressure (chamber II) to the side having higher osmotic pressure (chamber I).
Osmotic pressure is calculated using the formula:
$\pi = iCRT$
For potassium sulfate $(\mathrm{K_2SO_4})$, the van’t Hoff factor is $i = 3$.
So,
$\pi = 3 \times \left(\frac{50 \times 10^{-3}/174}{2}\right) \times 0.083 \times 300$
On solving,
$\pi = 0.0107\, \text{bar}$
When 0.25 moles of a non-volatile, non-ionizable solute was dissolved in 1 mole of a solvent the vapor pressure of solution was $x \%$ of vapor pressure of pure solvent. What is $x \%$ ?
50%
60%
70%
80%
Explanation:
Let the mole fraction of the solvent be $ x_1 $ and that of the solute be $ x_2 $.
Given:
Moles of solute $ n_2 = 0.25 $
Moles of solvent $ n_1 = 1 $
$ x_1 = \frac{n_1}{n_1 + n_2} = \frac{1}{1 + 0.25} = \frac{1}{1.25} = 0.8 $
$ x_2 = 1 - x_1 = 1 - 0.8 = 0.2 $
For a non-volatile solute, the vapor pressure of the solution is given by:
$ P_{\text{solution}} = x_1 P^0 $
where $ P^0 $ is the vapor pressure of the pure solvent.
$ \frac{P_{\text{solution}}}{P^0} \times 100 = x_1 \times 100 = 0.8 \times 100 = 80\% $
Final Answer:
$ x = 80\% $
Option D: 80%
At $27^{\circ} \mathrm{C}, 0.1 \mathrm{M}, 1 \mathrm{~L} \mathrm{~K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]$ aqueous solution and $0.1 \mathrm{M}, 1 \mathrm{~L} \mathrm{FeCl}_3$ aqueous solution are placed in a container separated by a semi permeable membrane AB . Assume complete dissociation of both the solutes. Which of the following statement is correct?
Blue color is formed on both sides.
Ionic solutes in aqueous solution can pass through semi-permeable membrane.
Solution on side ' y ' is hypotonic.
To cause the reverse flow of solvent during osmosis, external pressure (any value) should be applied to side ' $x$ '.
Explanation:
(A) A semi-permeable membrane (SPM) allows only solvent molecules (water) to pass through it. It does not allow ions to pass.
So, the solute ions from the two sides cannot come together. Therefore, no blue colour (due to mixing of ions) will appear on either side.
(B) Because ions cannot cross the SPM, only solvent molecules can move from one side to the other depending on osmotic pressure.
(C) A hypotonic solution is the one which has lower osmotic pressure (lower “effective” concentration of solute particles).
We use the NCERT relation for osmotic pressure:
As osmotic pressure $(\pi)=i C R T$
$ \begin{aligned} & \mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right] \longrightarrow 4 \mathrm{~K}^{+}+\mathrm{Fe}(\mathrm{CN})_6^{-4} \quad(\because i=5) \\ & \mathrm{FeCl}_3 \longrightarrow \mathrm{Fe}^{+3}+\mathrm{3Cl^-} \quad(\because i=4) \\ & \text { for side } x: \pi=5 \mathrm{CRT}=0.5 \mathrm{RT} \quad(\because c=0.1 \mathrm{~m}) \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, y: \pi=0.4 R T \quad(\because c=0.1 \mathrm{~m}) \end{aligned} $
effective concentration of $x=\dot{i} \times C=0.5 \mathrm{M}$
$ y=i \times C=0.4 \mathrm{~m} $
Since, effective concentration of $y$ is less than $x$. Hence, $y$ is hypotonic.
(D) In reverse osmosis, pressure is applied on the side having higher osmotic pressure (here, side $x$). But this applied external pressure must be greater than the osmotic pressure difference between the two sides.
i.e. P applied > $\left(\pi_x-\pi_y\right)$
Solution A is prepared by dissolving 1 g of a protein (molar mass = 50,000 g mol-1) in 0.5 L of water at 300 K. Its osmotic pressure is $x$ bar. Solution B is made by dissolving 2 g of the same protein in 1 L of water at 300 K. Osmotic pressure of solution B is $y$ bar. Entire solution of A is mixed with entire solution of B at the same temperature. The osmotic pressure of resultant solution is $z$ bar. $x$, $y$ and $z$ respectively are:
(R = 0.083 L bar mol-1 K-1)
$9.96 \times 10^{-4};\;9.96 \times 10^{-4};\;9.96 \times 10^{-4}$
$9.96 \times 10^{-4};\;9.96 \times 10^{-4};\;19.92 \times 10^{-4}$
$4.98 \times 10^{-4};\;4.98 \times 10^{-4};\;9.96 \times 10^{-4}$
$4.98 \times 10^{-4};\;4.98 \times 10^{-4};\;4.98 \times 10^{-4}$
Explanation:
For osmotic pressure, we use the NCERT relation:
$ \pi = C R T = \frac{n}{V}RT $
where
$\pi$ = osmotic pressure
$n$ = number of moles of solute
$V$ = volume of solution
$R = 0.083 \, \text{L bar mol}^{-1}\text{K}^{-1}$
$T = 300 \, \text{K}$
Protein molar mass is:
$ M = 50000 \, \text{g mol}^{-1} $
Now solve each part.
For solution A
Mass of protein $= 1 \, \text{g}$
Moles of protein:
$ n_A = \frac{1}{50000} = 2 \times 10^{-5} \, \text{mol} $
Volume:
$ V_A = 0.5 \, \text{L} $
So concentration:
$ C_A = \frac{2 \times 10^{-5}}{0.5} = 4 \times 10^{-5} \, \text{mol L}^{-1} $
Hence osmotic pressure:
$ x = \pi_A = C_A RT $
$ x = (4 \times 10^{-5})(0.083)(300) $
First calculate:
$ 0.083 \times 300 = 24.9 $
So,
$ x = 4 \times 10^{-5} \times 24.9 = 9.96 \times 10^{-4} \, \text{bar} $
For solution B
Mass of protein $= 2 \, \text{g}$
Moles:
$ n_B = \frac{2}{50000} = 4 \times 10^{-5} \, \text{mol} $
Volume:
$ V_B = 1 \, \text{L} $
Concentration:
$ C_B = \frac{4 \times 10^{-5}}{1} = 4 \times 10^{-5} \, \text{mol L}^{-1} $
Thus,
$ y = \pi_B = C_BRT $
$ y = (4 \times 10^{-5})(0.083)(300) $
$ y = 9.96 \times 10^{-4} \, \text{bar} $
After mixing A and B
Total moles of protein:
$ n_{\text{total}} = 2 \times 10^{-5} + 4 \times 10^{-5} = 6 \times 10^{-5} \, \text{mol} $
Total volume:
$ V_{\text{total}} = 0.5 + 1 = 1.5 \, \text{L} $
Concentration after mixing:
$ C_{\text{mix}} = \frac{6 \times 10^{-5}}{1.5} = 4 \times 10^{-5} \, \text{mol L}^{-1} $
Again,
$ z = \pi_{\text{mix}} = C_{\text{mix}}RT $
$ z = (4 \times 10^{-5})(0.083)(300) = 9.96 \times 10^{-4} \, \text{bar} $
So,
$ x = 9.96 \times 10^{-4} \, \text{bar}, \quad y = 9.96 \times 10^{-4} \, \text{bar}, \quad z = 9.96 \times 10^{-4} \, \text{bar} $
Therefore, the correct option is:
$ \boxed{\text{Option A}} $
19.5 g of fluoro acetic acid (molar mass = 78 g mol−1) is dissolved in 500 g of water at 298 K. The depression in the freezing point of water was 1∘C. What is $K_a$ of fluoro acetic acid?
(For water, $K_f = 1.86$ K kg mol−1). Assume molarity and molality to have same values.
$10^{-6}$
$4 \times 10^{-4}$
$3 \times 10^{-5}$
$3 \times 10^{-3}$
Explanation:
Given:
Mass of fluoroacetic acid $= 19.5 \, \text{g}$
Molar mass $= 78 \, \text{g mol}^{-1}$
Mass of water $= 500 \, \text{g} = 0.5 \, \text{kg}$
Depression in freezing point, $\Delta T_f = 1^\circ \text{C} = 1 \, \text{K}$
$K_f = 1.86 \, \text{K kg mol}^{-1}$
We have to find $K_a$ of fluoroacetic acid.
First, calculate the number of moles of acid:
$ n = \frac{19.5}{78} = 0.25 \, \text{mol} $
Now, molality of the solution:
$ m = \frac{0.25}{0.5} = 0.5 \, \text{mol kg}^{-1} $
For an electrolyte,
$ \Delta T_f = i K_f m $
So,
$ 1 = i \times 1.86 \times 0.5 $
$ 1 = i \times 0.93 $
$ i = \frac{1}{0.93} \approx 1.075 $
Fluoroacetic acid is a weak monoprotic acid:
$ \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- $
If degree of dissociation is $\alpha$, then for this type of acid:
$ i = 1 + \alpha $
Therefore,
$ \alpha = i - 1 = 1.075 - 1 = 0.075 $
Now use Ostwald’s dilution law:
$ K_a = \frac{C\alpha^2}{1-\alpha} $
Given in the question, assume molarity and molality are same, so
$ C = 0.5 $
Thus,
$ K_a = \frac{0.5 \times (0.075)^2}{1-0.075} $
$ = \frac{0.5 \times 0.005625}{0.925} $
$ = \frac{0.0028125}{0.925} \approx 3.04 \times 10^{-3} $
So,
$ K_a \approx 3 \times 10^{-3} $
Hence, the correct option is:
$ \boxed{3 \times 10^{-3}} $
Option D.
20 g hemoglobin in a 1 L aqueous solution $(\mathrm{A})$ at 300 K is separated from pure water by semi permeable membrane. At equilibrium the height of solution in a tube dipped in a solution (A) is found to be 80.0 mm higher than the tube dipped in water.
The molar mass of hemoglobin is $\_\_\_\_$ $\mathrm{kg} \mathrm{mol}^{-1}$. (Nearest integer)
(Given : $\mathrm{g}=10 \mathrm{~m} \mathrm{~s}^{-2}, \mathrm{R}=8.3 \mathrm{kPa} \mathrm{dm} \mathrm{K}^{-1} \mathrm{~mol}^{-1}$, density of solution $=1000 \mathrm{~kg} \mathrm{~m}^{-3}$ )
Explanation:
For a solution separated from pure water by a semipermeable membrane, at equilibrium the osmotic pressure is balanced by the hydrostatic pressure.
So,
$ \pi = h \rho g $
Given:
$h = 80.0 \, \text{mm} = 0.080 \, \text{m}$
$\rho = 1000 \, \text{kg m}^{-3}$
$g = 10 \, \text{m s}^{-2}$
Therefore,
$ \pi = 0.080 \times 1000 \times 10 $
$ \pi = 800 \, \text{Pa} = 0.8 \, \text{kPa} $
Now, using the osmotic pressure formula:
$ \pi = CRT $
where $C = \dfrac{n}{V}$
So,
$ \pi = \frac{w}{MV}RT $
Here,
$w = 20 \, \text{g}$
$M =$ molar mass
$V = 1 \, \text{L} = 1 \, \text{dm}^3$
$R = 8.3 \, \text{kPa dm}^3 \text{K}^{-1} \text{mol}^{-1}$
$T = 300 \, \text{K}$
$\pi = 0.8 \, \text{kPa}$
Substituting,
$ 0.8 = \frac{20}{M \times 1}\times 8.3 \times 300 $
$ 0.8 = \frac{20 \times 8.3 \times 300}{M} $
$ M = \frac{20 \times 8.3 \times 300}{0.8} $
$ M = 62250 \, \text{g mol}^{-1} $
Convert into $\text{kg mol}^{-1}$:
$ M = 62.25 \, \text{kg mol}^{-1} $
Nearest integer:
$ \boxed{62 \, \text{kg mol}^{-1}} $
A non-volatile, non-electrolyte solid solute when dissolved in 40 g of a solvent, the vapour pressure of the solvent decreased from 760 mm Hg to 750 mm Hg . If the same solution boils at 320 K , then the number of moles of the solvent present in the solution is $\_\_\_\_$ . (Nearest integer)
[Given: boiling point of the pure solvent $=319.5 \mathrm{~K}$,
$\mathrm{K}_{\mathrm{b}}$ of the solvent $=0.3 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ ]
Explanation:
For a non-volatile, non-electrolyte solute, first use relative lowering of vapour pressure:
$ \frac{p^0-p}{p^0}=x_2 $
where $x_2$ is the mole fraction of solute.
Given:
$ p^0=760\ \text{mm Hg}, \quad p=750\ \text{mm Hg} $
So,
$ x_2=\frac{760-750}{760}=\frac{10}{760}=\frac{1}{76} $
Now,
$ x_2=\frac{n_2}{n_1+n_2} $
where $n_1$ = moles of solvent and $n_2$ = moles of solute.
Hence,
$ \frac{n_2}{n_1+n_2}=\frac{1}{76} $
This gives
$ 76n_2=n_1+n_2 $
$ n_1=75n_2 $
Now use elevation in boiling point:
$ \Delta T_b = K_b m $
Given:
$ \Delta T_b=320-319.5=0.5\ \text{K} $
and
$ K_b=0.3\ \text{K kg mol}^{-1} $
So molality,
$ m=\frac{\Delta T_b}{K_b}=\frac{0.5}{0.3}=\frac{5}{3} $
But
$ m=\frac{n_2}{\text{mass of solvent in kg}} $
Mass of solvent $=40\text{ g}=0.040\text{ kg}$
Thus,
$ \frac{n_2}{0.040}=\frac{5}{3} $
$ n_2=\frac{5}{3}\times 0.040=0.0667\ \text{mol} $
Now,
$ n_1=75n_2=75\times 0.0667=5.00\ \text{mol} $
Therefore, the number of moles of solvent in the solution is
$ \boxed{5} $
Nearest integer: $\boxed{5}$
The freezing point depression of a 0.1 m aqueous solution of a monobasic weak acid HA is 0.20 °C. The dissociation constant for the acid is
Given: $K_f$(H2O) = 1.8 K kg mol−1, molality ≡ molarity
$1.90 \times 10^{-3}$
$1.38 \times 10^{-3}$
$1.1 \times 10^{-2}$
$1.89 \times 10^{-1}$
Explanation:
Freezing Point Depression:
$ \Delta T_f = i \times K_f \times m $
Given:
$ \Delta T_f = 0.2 \, \text{°C}, \quad K_f = 1.8 \, \text{K kg mol}^{-1}, \quad m = 0.1 \, \text{m} $
Substituting the given values:
$ 0.2 = i \times 1.8 \times 0.1 $
Solving for $i$:
$ i = \frac{0.2}{1.8 \times 0.1} = \frac{20}{18} = \frac{10}{9} $
Degree of Dissociation ($\alpha$):
For the reaction $\mathrm{HA} \rightleftharpoons \mathrm{H}^{+} + \mathrm{A}^{-}$:
$ i = 1 + \alpha $
Given $i = \frac{10}{9}$:
$ \frac{10}{9} = 1 + \alpha $
$ \alpha = \frac{1}{9} $
Dissociation Constant ($K_{eq}$):
$ \mathrm{K}_{eq} = \frac{[H^+][A^-]}{[HA]} $
At equilibrium:
$ [H^+] = [A^-] = \alpha \times C = \frac{1}{9} \times 0.1 $
$ [HA] = 0.1 \times (1 - \alpha) = 0.1 \times \left(1 - \frac{1}{9}\right) $
Substituting these into $K_{eq}$:
$ \mathrm{K}_{eq} = \frac{(0.1 \times \frac{1}{9})^2}{0.1 \times \left(1 - \frac{1}{9}\right)} $
Simplifying:
$ \mathrm{K}_{eq} = \frac{0.1 \times \left(\frac{1}{81}\right)}{0.1 \times \frac{8}{9}} = \frac{1}{720} $
Therefore:
$ \mathrm{K}_{eq} = 1.38 \times 10^{-3} $
Which of the following binary mixture does not show the behaviour of minimum boiling azeotropes?
$\text{CH}_3\text{OH} + \text{CHCl}_3$
$\text{C}_6\text{H}_5\text{OH} + \text{C}_6\text{H}_5\text{NH}_2$
$\text{H}_2\text{O} + \text{CH}_3\text{COC}_2\text{H}_5$
$\text{CS}_2 + \text{CH}_3\text{COCH}_3$
Explanation:
A binary mixture of $ \text{C}_6\text{H}_5\text{OH} $ and $ \text{C}_6\text{H}_5\text{NH}_2 $ exhibits negative deviation from Raoult's law. This means that the vapor pressure of the solution is lower than the vapor pressures of the pure components, $ \text{C}_6\text{H}_5\text{OH} $ and $ \text{C}_6\text{H}_5\text{NH}_2 $. Consequently, the boiling point of the solution is higher than the boiling points of the pure substances. Therefore, this mixture forms a maximum boiling azeotrope.
Liquid A and B form an ideal solution. The vapour pressures of pure liquids A and B are 350 and 750 mm Hg respectively at the same temperature. If $x_A$ and $x_B$ are the mole fraction of A and B in solution while $y_A$ and $y_B$ are the mole fraction of A and B in vapour phase, then,
$(x_A - y_A) < (x_B - y_B)$
$\frac{x_A}{x_B} = \frac{y_A}{y_B}$
$\frac{x_A}{x_B} < \frac{y_A}{y_B}$
$\frac{x_A}{x_B} > \frac{y_A}{y_B}$
Explanation:
Liquid A and B form an ideal solution. The vapor pressures of pure liquids A and B are 350 mm Hg and 750 mm Hg, respectively, at the same temperature. Here, $ x_A $ and $ x_B $ represent the mole fractions of A and B in the solution, and $ y_A $ and $ y_B $ are their mole fractions in the vapor phase.
Let’s begin by comparing the vapor pressures:
$ \mathrm{P}_{\mathrm{A}}^{\mathrm{o}} < \mathrm{P}_{\mathrm{B}}^{\mathrm{o}} $
$ \frac{\mathrm{P}_{\mathrm{A}}^{\mathrm{o}}}{\mathrm{P}_{\mathrm{B}}^{\mathrm{o}}} < 1 $
The relationship between the mole fractions in the vapor phase and the solution can be expressed as:
$ \frac{y_A}{y_B} = \frac{\mathrm{P}_{\mathrm{A}}^{\mathrm{o}}}{\mathrm{P}_{\mathrm{B}}^{\mathrm{o}}} \cdot \frac{x_A}{x_B} $
Since $\frac{\mathrm{P}_{\mathrm{A}}^{\mathrm{o}}}{\mathrm{P}_{\mathrm{B}}^{\mathrm{o}}} < 1$, it follows that:
$ \frac{\frac{y_A}{y_B}}{\frac{x_A}{x_B}} < 1 $
Which implies:
$ \frac{y_A}{y_B} < \frac{x_A}{x_B} $
This indicates that the mole fraction ratio of A to B in the vapor phase is less than that in the solution.
Match List - I with List - II.
| List - I | List - II |
|---|---|
| (A) Solution of chloroform and acetone | (I) Minimum boiling azeotrope |
| (B) Solution of ethanol and water | (II) Dimerizes |
| (C) Solution of benzene and toluene | (III) Maximum boiling azeotrope |
| (D) Solution of acetic acid in benzene | (IV) ΔVmix = 0 |
Choose the correct answer from the options given below :
(A)-(III), (B)-(I), (C)-(IV), (D)-(II)
(A)-(III), (B)-(IV), (C)-(I), (D)-(II)
(A)-(II), (B)-(I), (C)-(IV), (D)-(III)
(A)-(II), (B)-(IV), (C)-(I), (D)-(III)
Explanation:
To correctly match the items from List - I with those in List - II, let's analyze the characteristics of each solution:
(A) Solution of chloroform and acetone: This solution exhibits negative deviation from Raoult's law. It creates a maximum boiling azeotrope because the interactions between chloroform and acetone molecules are stronger than those in the pure components.
(B) Solution of ethanol and water: This solution shows positive deviation from Raoult's law, leading to the formation of a minimum boiling azeotrope. The interactions between ethanol and water molecules are weaker than those in their pure states.
(C) Solution of benzene and toluene: This combination forms an ideal solution where Raoult’s law is obeyed across all concentrations. Thus, the volume change upon mixing, denoted as $\Delta V_{\text{mix}}$, is zero.
(D) Solution of acetic acid in benzene: In this mixture, acetic acid tends to dimerize. The acetic acid molecules pair up, forming dimers, especially in non-polar solvents like benzene.
Based on these explanations, the correct matches are:
(A) - (III)
(B) - (I)
(C) - (IV)
(D) - (II)
Given below are two statements:
Statement (I) : Molal depression constant $\mathrm{K}_f$ is given by $\frac{\mathrm{M}_1 \mathrm{RT}_f}{\Delta \mathrm{~S}_{\mathrm{fus}}}$, where symbols have their usual meaning.
Statement (II) : $\mathrm{K}_f$ for benzene is less than the $\mathrm{K}_f$ for water.
In the light of the above statements, choose the most appropriate answer from the options given below :
Explanation:
Statement-I
Molar depression constant $\mathrm{k}_f=\frac{\mathrm{M}_1 \mathrm{RT}_f^2}{\Delta \mathrm{H}_{\text {fus }}}$
$\begin{aligned} & \mathrm{k}_f=\frac{\mathrm{M}_1 \mathrm{RT}_{\mathrm{f}}}{\left[\frac{\Delta \mathrm{H}_{\mathrm{fus}}}{\mathrm{~T}_{\mathrm{f}}}\right]} \\ & \mathrm{k}_f=\frac{\mathrm{M}_1 \mathrm{RT}_f}{\Delta \mathrm{~S}_{\text {fus }}} \end{aligned}$
Hence statement-I is correct
but $\mathrm{k}_{\mathrm{f}}$ for benzene $=5.12 \frac{{ }^{\circ} \mathrm{C}}{\text { molal }}$
$\mathrm{k}_{\mathrm{f}}$ for water $=1.86 \frac{{ }^{\circ} \mathrm{C}}{\text { molal }}$ Hence statement- II is incorrect
$X Y$ is the membrane/partition between two chambers 1 and 2 containing sugar solutions of concentration $c_1$ and $c_2\left(c_1>c_2\right) \mathrm{mol} \mathrm{L}^{-1}$. For the reverse osmosis to take place identify the correct condition.
(Here $p_1$ and $p_2$ are pressures applied on chamber 1 and 2 ).
A. Membrane/Partition : Cellophane, $\mathrm{p}_1>\pi$
B. Membrane/Partition : Porous, $\mathrm{p}_2>\pi$
C. Membrane/Partition : Parchment paper, $p_1>\pi$
D. Membrane/Partition : Cellophane, $\mathrm{p}_2>\pi$
Choose the correct answer from the option given below:
Explanation:
Normal osmosis occurs from (2) to (1)
For reverse osmosis from (1) to (2)
Pressure : $\mathrm{P}_1>\pi$
$\therefore$ Answer [A & C] only
Explanation:
Boiling Point Elevation Formula:
$ \Delta T_b = i_1 \cdot m_1 \cdot K_b + i_2 \cdot m_2 \cdot K_b $
where:
$i_1$ and $i_2$ are the van't Hoff factors for ethylene glycol and glucose, respectively (both are 1 since they do not dissociate in solution).
$m_1$ and $m_2$ are the molalities of ethylene glycol and glucose, respectively.
$K_b$ is the ebullioscopic constant of water ($0.52 \, \text{K kg mol}^{-1}$).
Calculate Molality:
Each solute has 2 moles dissolved in 500 grams of water ($0.5 \, \text{kg}$):
$ m_1 = \frac{2 \, \text{moles}}{0.5 \, \text{kg}} = 4 \, \text{mol kg}^{-1} $
$ m_2 = \frac{2 \, \text{moles}}{0.5 \, \text{kg}} = 4 \, \text{mol kg}^{-1} $
Substitute into the Formula:
$ \Delta T_b = 1 \cdot 4 \cdot 0.52 + 1 \cdot 4 \cdot 0.52 = 4.16 $
Determine Boiling Point of Solution:
The normal boiling point of water is $373.16 \, \text{K}$.
Add the boiling point elevation to the normal boiling point:
$ T_b (\text{solution}) = 373.16 + 4.16 = 377.3 \, \text{K} $
Thus, the boiling point of the resulting solution is $377.3 \, \text{K}$.
Which of the following properties will change when system containing solution 1 will become solution 2 ?
Explanation:
Both solutions contain the same composition, specifically 1 mole of 'x' in 1 liter of water. This means all intensive properties, such as concentration and density, will remain unchanged. However, because the total quantity of solution is greater in Solution 1 compared to Solution 2, the extensive properties will differ. Consequently, Gibbs free energy will change as it is an extensive property.
' $x$ ' g of NaCl is added to water in a beaker with a lid. The temperature of the system is raised from $1^{\circ} \mathrm{C}$ to $25^{\circ} \mathrm{C}$. Which out of the following plots, is best suited for the change in the molarity $(\mathrm{M})$ of the solution with respect to temperature ?
[Consider the solubility of NaCl remains unchanged over the temperature range]
Explanation:
When $x$ grams of NaCl are added to water in a beaker and the temperature is increased from $1^{\circ} \mathrm{C}$ to $25^{\circ} \mathrm{C}$, the molarity of the solution changes due to the volumetric changes of water.
Temperature from $1^{\circ} \mathrm{C}$ to $4^{\circ} \mathrm{C}$:
Water is densest at $4^{\circ} \mathrm{C}$.
As the temperature increases from $1^{\circ} \mathrm{C}$ to $4^{\circ} \mathrm{C}$, the water volume decreases due to increased density.
This decrease in volume results in an increase in molarity because molarity is inversely proportional to the solution's volume.
Temperature from $4^{\circ} \mathrm{C}$ to $25^{\circ} \mathrm{C}$:
Beyond $4^{\circ} \mathrm{C}$, water expands with an increase in temperature.
Therefore, as temperature rises to $25^{\circ} \mathrm{C}$, the volume of the water increases.
The dilution leads to a decrease in molarity, since molarity is inversely proportional to volume.
Thus, the molarity first increases as temperature rises to $4^{\circ} \mathrm{C}$, but then decreases as it continues to $25^{\circ} \mathrm{C}$. The graphical representation of this relationship would exhibit an initial increase in molarity, followed by a decrease, correlating with changes in the volume of water due to temperature variations.
A solution is made by mixing one mole of volatile liquid $A$ with 3 moles of volatile liquid $B$. The vapour pressure of pure A is 200 mm Hg and that of the solution is 500 mm Hg . The vapour pressure of pure B and the least volatile component of the solution, respectively, are:
Explanation:
Given:
1 mole of volatile liquid A
3 moles of volatile liquid B
Vapor pressure of pure A, $ P_A^o = 200 $ mm Hg
Vapor pressure of the solution, $ P_{S} = 500 $ mm Hg
We apply Raoult's law, which states:
$ P_{S} = P_A^o \cdot X_A + P_B^o \cdot X_B $
Where:
$ X_A $ is the mole fraction of A
$ X_B $ is the mole fraction of B
$ P_B^o $ is the vapor pressure of pure liquid B
Calculate the mole fractions:
$ X_A = \frac{1}{1+3} = \frac{1}{4} $
$ X_B = \frac{3}{1+3} = \frac{3}{4} $
Plug these into the equation:
$ 500 = 200 \times \frac{1}{4} + P_B^o \times \frac{3}{4} $
Simplifying:
$ 500 = 50 + \frac{3}{4} P_B^o $
Subtract 50 from both sides:
$ 450 = \frac{3}{4} P_B^o $
Multiply both sides by $\frac{4}{3}$ to solve for $P_B^o$:
$ P_B^o = 600 \, \text{mm Hg} $
Since $ P_A^o < P_B^o $, liquid A is the least volatile component.
In conclusion:
The vapor pressure of pure B, $ P_B^o $, is 600 mm Hg.
The least volatile component is A.
Which of the following graph correctly represents the plots of $\mathrm{K}_{\mathrm{H}}$ at 1 bar for gases in water versus temperature?
Explanation:
As temperature increases solubility first decrease then increase hence $\mathrm{K}_{\mathrm{H}}$ first increase than decrease also at moderate temperature $\mathrm{K}_{\mathrm{H}}$ value $\mathrm{He}>\mathrm{N}_2>$ $\mathrm{CH}_4$.
Given below are two statements :
Statement (I): NaCl is added to the ice at 0°C, present in the ice cream box to prevent the melting of ice cream.
Statement (II): On addition of NaCl to ice at 0°C, there is a depression in freezing point.
In the light of the above statements, choose the correct answer from the options given below :
Both Statement I and Statement II are false
Statement I is true but Statement II is false
Both Statement I and Statement II are true
Statement I is false but Statement II is true
Explanation:
Statement I : Correct
NaCl addition to ice causes preventing the melting of ice. On adding NaCl to ice, freezing point lowers. This creates a colder mixture, preventing the ice cream from melting.
Melting point of ice is 0$^\circ$C. When only ice is used to make ice cream, at 0$^\circ$C ice starts melting by absorbing the energy from its environment in the form of heat. Addition of salt to ice while making the cream lowers the freezing point of the ice, allowing it to reach a colder temperature and thus the ice cream mixture freezes properly, So, the salt causes ice to melt at a lower temperature than pure ice.
Statement II : Correct
Decrease in freezing point while addition of NaCl to ice at 0$^\circ$C is due to the colligative property depression in freezing point.
So, both the statements are correct.
Both statement I and statement II are true.
1.24 g of AX2 (molar mass 124 g mol−1) is dissolved in 1 kg of water to form a solution with boiling point of 100.015°C, while 25.4 g of AY2 (molar mass 250 g mol−1) in 2 kg of water constitutes a solution with a boiling point of 100.0260°C.
Kb(H2O) = 0.52 kg mol−1
Which of the following is correct?
AX2 and AY2 (both) are completely unionised.
AX2 and AY2 (both) are fully ionised.
AX2 is completely unionised while AY2 is fully ionised.
AX2 is fully ionised while AY2 is completely unionised.
Explanation:
Mass of $A x_2=1.24 \mathrm{~g}$ (solute)
Molarmass of $A X_2=124 \mathrm{~g} \mathrm{~mol}^{-1}$
Mass of water $=1 \mathrm{~kg}$ (solvent.)
Boiling point of water $=100^{\circ} \mathrm{C}$
Boiling point of water after adding solute $A X_2=100.0156^{\circ} \mathrm{C}$
Mass of $A Y_2=25.4 \mathrm{~g}$ (solute)
Molarmass of $A Y_2=250 \mathrm{~g~mol}^{-1}$
Mass of water $=2 \mathrm{~kg}$ (Solvent)
Boiling point of water $\mp 100^{\circ} \mathrm{C}$
Boiling point of water after adding solute $A y_2=100.0260^{\circ} \mathrm{C}$
$\mathrm{K}_{\mathrm{b}}\left(\mathrm{H}_2 \mathrm{O}\right)=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}{ }^{-1}=0.520^{\circ} \mathrm{C} \mathrm{kg} \mathrm{mol}^{-1}$
The ionisation of $A x_2$ and $A Y_2$ can be determined by calculating Van't Hoff factor.
The phenomenon given in the question is elevation in boiling point. The boiling point of solvent increases when another compound (solute) is added to it.
Relation: $\Delta T_b=K_b \cdot i \cdot m$
$\Delta T_b=T_b-T_b^0$
$\Delta T_b \rightarrow$ Boiling point elevation
$T_b \rightarrow$ Boiling point of solution (solvent + solute)
$T_b^0 \rightarrow$ Boiling point of solvent
$K_b \rightarrow$ Molal elevation constant
$i \rightarrow$ Van't Hoff factor
$m \rightarrow \text { Molality }=\frac{\text { Number of moles }}{\mathrm{kg} \text { of solvent }}, \text { Moles }=\frac{\text { Mass }}{\text { Molarmass }}$
For $A X_2$ and $A Y_2$ (solute), Van't Hoff factor represents how many particles a solute dissociates into when dissolved in a solvent (water).
For non-electrolytes, $i=1$
$\begin{aligned} &\begin{aligned} & A X_2: \\ & \text { Moles }=\frac{\text { mass }}{\text { molarmass }}=\frac{1.249}{124 \mathrm{~g~mol^{-1}}}=0.01 \mathrm{~mol} \\ & \text { Molality }=\frac{\text { Moles }}{k_g \text { of solvent }}=\frac{0.01 \mathrm{~mol}}{1 \mathrm{~kg}}=0.01 \mathrm{~mol} \mathrm{~kg}^{-1} \\ & \Delta T_b=K_b \times i \times m \\ & i=\frac{\Delta T_b}{k_b \times m} \\ &=\frac{T_b-T_b^0}{K_b \times m} \end{aligned}\\ &\text { Substitute values as., } \end{aligned}$
$\begin{aligned} i & =\frac{\left(100.0156^{\circ} \mathrm{C}-100^{\circ} \mathrm{C}\right)}{0.52^{\circ} \mathrm{C} \mathrm{~kg} \mathrm{~mol}^{-1} \times 0.01 \mathrm{~mol} \mathrm{~kg}^{-1}} \\ & =\frac{0.0156^\circ \mathrm{C}}{0.52 \times 0.01{ }^{\circ} \mathrm{C}}=3 \end{aligned}$
$i=3$ means, there are 3 particles in solution after $A X_2$ dissolved in water.
$A x_2 \rightarrow 1 A^{+}+2 X^{-} \quad(1+2=3)$
$\begin{aligned} &A x_2 \text { is completely ionised. }\\ &A Y_2: \end{aligned}$
$\begin{aligned} &\begin{aligned} \text { Moles } & =\frac{\text { Mass }}{\text { Molarmass }}=\frac{25.4 \mathrm{~g}}{250 \mathrm{g~mol}}=0.1016 \mathrm{~mol} \\ \text { Molality } & =\frac{\text { Moles }}{\mathrm{kgof}^{-1} \text { solvent }}=\frac{0.1016 \mathrm{~mol}}{2 \mathrm{~kg}}=0.050 .8 \mathrm{~mol} \mathrm{~kg}^{-1} \\ \Delta T_b & =k_b \times \mathrm{l}^2 \times \mathrm{m} \\ i & =\frac{\Delta T_b}{k_b \times m} \\ & =\frac{T_b-T_b^0}{k_b \times m} \end{aligned}\\ &\text { Substitute values as, } \end{aligned}$
$i = {{(100.0260^\circ C - 100^\circ C)} \over {0.52^\circ C\,kg\,mo{l^{ - 1}} \times 0.0508\,mol\,k{g^{ - 1}}}}$
$=\frac{0.0260 ^\circ \mathrm{C}}{0.52 \times 0.0508 ^\circ \mathrm{C}}$
$= 0.98$
$\approx 1$
$i=1$ means $A Y_2$ is completely unionised.
$A y_2$ not give ionised particles when dissolved in water.
So, $A X_2$ is completely ionised and $A Y_2$ is completely unionised.
Answer: Option 4) $A x_2$ is fully ionised, $A Y_2$ is completely unionised.
Assume a living cell with 0.9% (w/w) of glucose solution (aqueous). This cell is immersed in another solution having equal mole fraction of glucose and water.
(Consider the data upto first decimal place only)
The cell will :
Explanation:
Living cell $=0.9 \mathrm{gm}$ in 100 gm of solution $\% \mathrm{w} / \mathrm{w}=0.9$
Solution is have equal moles of glucose and water $=0.5$
Weight of solution $=0.5 \times 180+0.5 \times 18=99 \mathrm{gm}$ $\% \mathrm{w} / \mathrm{w} \simeq 90 \%$
Concentrated solution $=$ Cell will shrink
What is the freezing point depression constant of a solvent, 50 g of which contain 1 g non volatile solute (molar mass $256 \mathrm{~g} \mathrm{~mol}^{-1}$ ) and the decrease in freezing point is 0.40 K ?
Explanation:
To find the freezing point depression constant ($ K_f $) of the solvent, we use the formula for freezing point depression:
$ \Delta T_f = K_f \cdot m $
Given:
The decrease in freezing point $\Delta T_f$ is 0.40 K.
The mass of the solute is 1 g and its molar mass is 256 g/mol.
The mass of the solvent is 50 g (or 0.050 kg).
First, calculate the molality ($ m $):
Molality is defined as the moles of solute per kilogram of solvent.
Calculate moles of solute:
$ \text{Moles of solute} = \frac{1 \, \text{g}}{256 \, \text{g/mol}} = \frac{1}{256} \, \text{mol} $
Calculate molality ($ m $):
$ m = \frac{\frac{1}{256} \, \text{mol}}{0.050 \, \text{kg}} = \frac{1}{256 \times 0.050} \, \text{mol/kg} $
Now, substitute into the formula to find $ K_f $:
$ 0.4 = K_f \cdot \frac{1}{256 \times 0.050} $
Solving for $ K_f $:
$ K_f = 0.4 \cdot 256 \times 0.050 = 5.12 \, \text{K kg/mol} $
Thus, the freezing point depression constant of the solvent is $ 5.12 \, \text{K kg/mol} $.
Consider the given plots of vapour pressure (VP) vs temperature(T/K). Which amongst the following options is correct graphical representation showing $\Delta \mathrm{T}_{\mathrm{f}}$, depression in the freezing point of a solvent in a solution?
Explanation:
On adding non-volatile solute in a solvent, the freezing point of solution decreases.
$\mathrm{T}_{\mathrm{f}}<\mathrm{T}_{\mathrm{f}}^0$
F.P. of solution $<$ F.P. of pure solvent
Also V.P. of solution decreases on adding nonvolatile solute in a solvent.
When a non-volatile solute is added to the solvent, the vapour pressure of the solvent decreases by 10 mm of Hg . The mole fraction of the solute in the solution is 0.2 . What would be the mole fraction of the solvent if decrease in vapour pressure is 20 mm of Hg ?
Explanation:
When a non-volatile solute is added to a solvent, it causes the vapour pressure of the solvent to decrease. In this scenario, when the vapour pressure decreases by 10 mm of Hg, the mole fraction of the solute in the solution is 0.2.
By understanding the relationship between vapour pressure change and mole fraction, we see that:
The change in vapour pressure ($P^{\circ} - P$) is directly proportional to the mole fraction of the solute ($X_{\text{solute}}$).
Therefore, if a 10 mm of Hg decrease corresponds to a mole fraction of 0.2, then a 20 mm of Hg decrease would correspond to a mole fraction of 0.4.
To find the mole fraction of the solvent ($X_{\text{solvent}}$), we use the formula:
$ X_{\text{solvent}} = 1 - X_{\text{solute}} $
Substituting the value we found:
$ X_{\text{solvent}} = 1 - 0.4 = 0.6 $
Thus, when the vapour pressure decreases by 20 mm of Hg, the mole fraction of the solvent is 0.6.
Consider a binary solution of two volatile liquid components 1 and $2 . x_1$ and $y_1$ are the mole fractions of component 1 in liquid and vapour phase, respectively. The slope and intercept of the linear plot of $\frac{1}{x_1}$ vs $\frac{1}{y_1}$ are given respectively as :
Explanation:
For a binary solution of two volatile liquid components labeled 1 and 2, let $ x_1 $ and $ y_1 $ represent the mole fractions of component 1 in the liquid and vapor phases, respectively. The linear relationship between the inverse of these mole fractions is plotted as $\frac{1}{x_1}$ versus $\frac{1}{y_1}$.
To derive the slope and intercept of this linear plot, consider the following calculations:
Using Raoult's Law for a Liquid Solution:
For a liquid solution with volatile components 1 and 2:
$ \mathrm{P}_1 = \mathrm{P}_{\mathrm{T}} \cdot y_1 = \mathrm{P}_1^{\mathrm{o}} \cdot x_1 $
Therefore:
$ \frac{\mathrm{P}_{\mathrm{T}}}{x_1} = \frac{\mathrm{P}_1^{\mathrm{o}}}{y_1} $
Rearranging the Equation:
By substituting and rearranging, we have:
$ \frac{\mathrm{P}_2^{\mathrm{o}} + x_1(\mathrm{P}_1^{\mathrm{o}} - \mathrm{P}_2^{\mathrm{o}})}{x_1} = \frac{\mathrm{P}_1^{\mathrm{o}}}{y_1} $
Simplifying further:
$ \frac{\mathrm{P}_2^{\mathrm{o}}}{x_1} + (\mathrm{P}_1^{\mathrm{o}} - \mathrm{P}_2^{\mathrm{o}}) = \frac{\mathrm{P}_1^{\mathrm{o}}}{y_1} $
Expressing $\frac{1}{x_1}$:
Solving for $\frac{1}{x_1}$, we obtain:
$ \frac{1}{x_1} = \left(\frac{\mathrm{P}_1^{\mathrm{o}}}{\mathrm{P}_2^{\mathrm{o}}}\right)\left(\frac{1}{y_1}\right) + \left(\frac{\mathrm{P}_2^{\mathrm{o}} - \mathrm{P}_1^{\mathrm{o}}}{\mathrm{P}_2^{\mathrm{o}}}\right) $
Determining the Slope and Intercept:
The slope of the line is:
$ \text{Slope} = \frac{\mathrm{P}_1^{\mathrm{o}}}{\mathrm{P}_2^{\mathrm{o}}} $
The intercept of the line is:
$ \text{Intercept} = \frac{\mathrm{P}_2^{\mathrm{o}} - \mathrm{P}_1^{\mathrm{o}}}{\mathrm{P}_2^{\mathrm{o}}} $
In summary, for the plot of $\frac{1}{x_1}$ against $\frac{1}{y_1}$, the slope is $\frac{\mathrm{P}_1^{\mathrm{o}}}{\mathrm{P}_2^{\mathrm{o}}}$ and the intercept is $\frac{\mathrm{P}_2^{\mathrm{o}} - \mathrm{P}_1^{\mathrm{o}}}{\mathrm{P}_2^{\mathrm{o}}}$.
Arrange the following solutions in order of their increasing boiling points.
(i) $10^{-4} \mathrm{M} \mathrm{NaCl}$
(ii) $10^{-4} \mathrm{M}$ Urea
(iii) $10^{-3} \mathrm{M} \mathrm{NaCl}$
(iv) $10^{-2} \mathrm{M} \mathrm{NaCl}$
Explanation:
Step 1: Identify the van’t Hoff factor ($i$) for each solute
NaCl dissociates (ideally) into two ions:
$ \mathrm{NaCl} \;\rightarrow\; \mathrm{Na^+} + \mathrm{Cl^-}, $
so $i \approx 2.$
Urea ($\mathrm{CH_4N_2O}$) is a non‐electrolyte (does not dissociate), so $i = 1.$
Step 2: Effective molar concentration of particles
The total particle concentration for each solution is approximately $(i \times \text{molarity})$.
(i) $10^{-4}\,M$ NaCl
$ \text{Effective concentration} \;=\; 2 \times 10^{-4} = 2 \times 10^{-4}. $
(ii) $10^{-4}\,M$ Urea
$ \text{Effective concentration} \;=\; 1 \times 10^{-4} = 1 \times 10^{-4}. $
(iii) $10^{-3}\,M$ NaCl
$ \text{Effective concentration} \;=\; 2 \times 10^{-3} = 2 \times 10^{-3}. $
(iv) $10^{-2}\,M$ NaCl
$ \text{Effective concentration} \;=\; 2 \times 10^{-2} = 2 \times 10^{-2}. $
Step 3: Compare to rank the boiling points
A larger total particle concentration (and hence larger colligative effect) corresponds to a higher boiling point. Arrange from lowest to highest:
Lowest: $10^{-4}\,M$ Urea $\bigl[1 \times 10^{-4}\bigr]$
Next: $10^{-4}\,M$ NaCl $\bigl[2 \times 10^{-4}\bigr]$
Next: $10^{-3}\,M$ NaCl $\bigl[2 \times 10^{-3}\bigr]$
Highest: $10^{-2}\,M$ NaCl $\bigl[2 \times 10^{-2}\bigr]$
Hence, in the format $(\text{ii}) < (\text{i}) < (\text{iii}) < (\text{iv})$.
Final Answer
$ \boxed{\text{(ii) } < \text{(i) } < \text{(iii) } < \text{(iv)}} \quad \text{(Option B)} $
Sea water, which can be considered as a 6 molar $(6 \mathrm{M})$ solution of NaCl , has a density of $2 \mathrm{~g} \mathrm{~mL}^{-1}$. The concentration of dissolved oxygen $\left(\mathrm{O}_2\right)$ in sea water is 5.8 ppm . Then the concentration of dissolved oxygen $\left(\mathrm{O}_2\right)$ in sea water, is $x \times 10^{-4} \mathrm{~m}$.
$x=$ ___________. (Nearest integer)
Given: Molar mass of NaCl is $58.5 \mathrm{~g} \mathrm{~mol}^{-1}$
Molar mass of $\mathrm{O}_2$ is $32 \mathrm{~g} \mathrm{~mol}^{-1}$
Explanation:
Sea water is 6 Molar in NaCl , So 1000 ml of sea water contains 6 mol of NaCl .
$\begin{aligned} & \text { mass of solution }=\text { Volume } \times \text { density } \\ & =1000 \times 2 \\ & \text { mass of solution }=2000 \mathrm{~g} \\ & \mathrm{ppm}=\frac{\text { mass of } \mathrm{O}_2}{2000} \times 10^6 \\ & \text { mass of } \mathrm{O}_2=5.8 \times 2 \times 10^{-3} \\ & \quad=1.16 \times 10^{-2} \mathrm{~g} \\ & \text { molality for } \mathrm{O}_2=\frac{1.16 \times 10^{-2} / 32}{(2000-6 \times 58.5)} \times 1000 \end{aligned}$
$\begin{aligned} &\begin{aligned} & =\frac{1.16 \times 10}{32 \times 1649} \\ & =0.000219 \\ & =2.19 \times 10^{-4} \end{aligned}\\ &\text { Correct answer } \Rightarrow 2 \end{aligned}$
When 1 g each of compounds AB and $\mathrm{AB}_2$ are dissolved in 15 g of water separately, they increased the boiling point of water by 2.7 K and 1.5 K respectively. The atomic mass of A (in $a m u$ ) is____________ $\times 10^{-1}$ (Nearest integer)
(Given : Molal boiling point elevation constant is $0.5 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ )
Explanation:
For AB
$\begin{aligned} & \Delta \mathrm{T}_{\mathrm{b}}=2.7 \mathrm{~K} \\ & 2.7=1 \times 0.5 \times \mathrm{m} \\ & \mathrm{~m}=\frac{27}{5} \end{aligned}$
Let molar mass of $A B=x$.
So $\frac{1 / x}{15} \times 1000=\frac{27}{5}$
$x=12.34$
For $\mathrm{AB}_2$
$\begin{aligned} & \Delta \mathrm{T}_{\mathrm{b}}=1.5 \mathrm{~K} \\ & 1.5=1 \times 0.5 \times \mathrm{m} \\ & \mathrm{~m}=3 \end{aligned}$
Let molar mass of $\mathrm{AB}_2=\mathrm{y}$
So $\frac{1 / \mathrm{y}}{15} \times 1000=3$
$\begin{aligned} & y=\frac{1000}{45} \\ & y=22.22 \end{aligned}$
Now let a and b be atomic masses of A and B respectively, then
$\begin{aligned} & \mathrm{A}+\mathrm{b}=12.34 \quad\text{...... (i)}\\ & \mathrm{~A}+2 \mathrm{~b}=22.22 \quad\text{...... (ii)}\\ & \mathrm{~B}=22.22-12.34=9.88 \end{aligned}$
Now $\mathrm{a}=12.34-9.88=2.46$
$=24.6 \times 10^{-1}=25 \times 10^{-1}$
If A2B is 30% ionised in an aqueous solution, then the value of van't Hoff factor (i) is _______ × 10−1.
Explanation:
Percent ionisation of $A_2B=30\%$
The dissociation of $A_2B$ in aqueous solution can be represented as
$A_2B \rightarrow 2A^+ + B^{2-}$
1 mole of $A_2B$ produces 2 moles of $A^+$ and 1 mole of $B^{2-}$ ions.
The total number of moles of ions produced (n) from the dissociation is
$n=2+1=3$
The degree of dissociation given that $A_2B$ is 30% ionised, the degree of dissociation $(\lambda)$ or degree of ionisation
$\lambda=\frac{30}{100}=0.3$
Van't Hoff factor ($i$) can be calculated using the formula,
$i=1+(n-1)\lambda$
Substitute the values of $n$ and $\lambda$ as,
$i = 1 + (3 - 1) \times 0.3$
$ = 1 + 2 \times 0.3$
$ = 1 + 0.6$
$ = 1.6$
$ = 16 \times {10^{ - 1}}$
At 300 K , an ideal dilute solution of a macromolecule exerts osmotic pressure that is expressed in terms of the height $(h)$ of the solution (density $=1.00 \mathrm{~g} \mathrm{~cm}^{-3}$ ) where $h$ is equal to 2.00 cm . If the concentration of the dilute solution of the macromolecule is $2.00 \mathrm{~g} \mathrm{dm}^{-3}$, the molar mass of the macromolecule is calculated to be $\boldsymbol{X} \times 10^4 \mathrm{~g} \mathrm{~mol}^{-1}$. The value of $\boldsymbol{X}$ is __________.
Use: Universal gas constant $(R)=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ and acceleration due to gravity $(g)=10 \mathrm{~m} \mathrm{~s}^{-2}$
Explanation:
Density of the solution:
$ \text{Density} = 1.00 \, \text{g cm}^{-3} = 1000 \, \text{kg m}^{-3} $
Height of the solution column ($h$):
$ h = 2 \, \text{cm} = 2 \times 10^{-2} \, \text{m} $
Acceleration due to gravity ($g$):
$ g = 10 \, \text{m s}^{-2} $
Calculating osmotic pressure ($\pi$):
Osmotic pressure is given by the equation:
$ \pi = h \cdot \rho \cdot g $
Substituting the known values:
$ \pi = 2 \times 10^{-2} \times 1000 \times 10 = 200 \, \text{N m}^{-2} $
Relating osmotic pressure to concentration and molar mass:
The equation for osmotic pressure in terms of concentration and molar mass is:
$ \pi = cRT $
where $c = \frac{2000}{M}$ as the concentration in g/dm$^3$ needs to be converted to mol/dm$^3$ by dividing by the molar mass $M$. Incorporating the given temperature and universal gas constant, we have:
$ 200 = \left(\frac{2000}{M}\right) \cdot 8.3 \cdot 300 $
Solving for the molar mass $M$:
Rearrange and solve the equation to find $M$:
$ M = 24900 \, \text{g mol}^{-1} = 2.49 \times 10^4 \, \text{g mol}^{-1} $
Thus, the value of $X$ is $2.49$.
$0.05 \mathrm{M} \mathrm{~CuSO}_4$ when treated with $0.01 \mathrm{M} \mathrm{~K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ gives green colour solution of $\mathrm{Cu}_2 \mathrm{Cr}_2 \mathrm{O}_7$. The two solutions are separated as shown below : [SPM : Semi Permeable Membrane]
Due to osmosis :
Explanation:
Osmosis leads to the net movement of water molecule from low osmotic pressure to high osmotic pressure. Since the iM value of $0.05 \mathrm{~M} \mathrm{~CuSO}_4$ solution is higher hence it has higher osmotic pressure so the water moves towards $\mathrm{CuSO}_4$ solution leads to drop of its molarity. The solute molecules do not cross S.P.M. via osmosis.