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 $
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.
Explanation:
| Salt | Values of i (for different conc. of a Salt) | ||
|---|---|---|---|
| NaCl | 0.1 M | 0.01 M | 0.001 M |
| 1.87 | 1.94 | 1.94 | |
The van 't Hoff factor (i) is used to describe the number of particles a solute formula unit produces in a solution. For an electrolyte like $\mathrm{NaCl}$, which dissociates completely in very dilute solutions, the theoretical value of $i$ is approximately 2, since $\mathrm{NaCl}$ dissociates into $\mathrm{Na}^+$ and $\mathrm{Cl}^-$ ions.
In real scenarios, as the concentration of the solution decreases (making the solution more dilute), the interaction between the ions decreases, allowing more complete dissociation. Therefore, for practical purposes, the van 't Hoff factor $i$ approaches its theoretical maximum value as concentration decreases. Thus, for $\mathrm{NaCl}$ solutions of concentrations $0.1 \mathrm{M}$, $0.01 \mathrm{M}$, and $0.001 \mathrm{M}$, the fact that $\mathrm{NaCl}$ dissociates more completely in more dilute solutions implies that $i$ increases with decreasing concentration.
Hence, the order of $i$ based on the concentration would be $\mathrm{i}_{\mathrm{A}} < \mathrm{i}_{\mathrm{B}} < \mathrm{i}_{\mathrm{C}}$, as concentration $0.1 \mathrm{M} > 0.01 \mathrm{M} > 0.001 \mathrm{M}$, respectively. Thus, option B correctly describes the order of the van 't Hoff factors for these solutions.
Identify the mixture that shows positive deviations from Raoult's Law
Explanation:
$\left(\mathrm{CH}_3\right)_2 \mathrm{CO}+\mathrm{CS}_2$ Exibits positive deviations from Raoult's Law
The solution from the following with highest depression in freezing point/lowest freezing point is
Explanation:
$\Delta \mathrm{T}_{\mathrm{f}}$ is maximum when $\mathrm{i} \times \mathrm{m}$ is maximum.
1) $\mathrm{m}_1=\frac{180}{60}=3, \mathrm{i}=1+\alpha$
Hence
$\Delta \mathrm{T}_{\mathrm{f}}=(1+\alpha) \cdot \mathrm{k}_{\mathrm{f}}=3 \times 1.86=5.58^{\circ} \mathrm{C}(\alpha<<1)$
2) $\mathrm{m}_2=\frac{180}{60}=3, \mathrm{i}=0.5, \Delta \mathrm{T}_{\mathrm{f}}=\frac{3}{2} \times \mathrm{k}_{\mathrm{f}}{ }^{\prime}=7.68^{\circ} \mathrm{C}$
3) $\mathrm{m}_3=\frac{180}{122}=1.48, \mathrm{i}=0.5, \Delta \mathrm{T}_{\mathrm{f}}=\frac{1.48}{2} \times \mathrm{k}_{\mathrm{f}}{ }^{\prime}=3.8^{\circ} \mathrm{C}$
4) $\mathrm{m}_4=\frac{180}{180}=1, \mathrm{i}=1, \Delta \mathrm{T}_{\mathrm{f}}=1 \cdot \mathrm{k}_{\mathrm{f}}{ }^{\prime}=1.86^{\circ} \mathrm{C}$
As per NCERT, $\mathrm{k}_{\mathrm{f}}{ }^{\prime}\left(\mathrm{H}_2 \mathrm{O}\right)=1.86 \mathrm{~k} \cdot \mathrm{~kg} \mathrm{~mol}^{-1}$
$\mathrm{k}_{\mathrm{f}}{ }^{\prime}(\text { Benzene })=5.12 \mathrm{~k} \cdot \mathrm{~kg} \mathrm{~mol}^{-1}$
What happens to freezing point of benzene when small quantity of napthalene is added to benzene?
Explanation:
When a small quantity of naphthalene is added to benzene, the freezing point of benzene decreases. This phenomenon is due to the colligative property known as freezing point depression. The addition of a solute, such as naphthalene, disrupts the orderly arrangement of solvent molecules in the solid phase, thereby lowering the freezing point.
Mathematically, the decrease in freezing point ($\Delta T_f$) can be represented by the equation:
$ \Delta T_f = i \cdot K_f \cdot m $
where:
$i$ is the van't Hoff factor (which is 1 for naphthalene as it does not dissociate in solution),
$K_f$ is the cryoscopic constant of the solvent (benzene),
$m$ is the molality of the solution.
Thus, the correct option is:
Option B: Decreases
A solution of two miscible liquids showing negative deviation from Raoult's law will have :
Explanation:
- Negative deviation means the intermolecular forces of attraction between the molecules of the two liquids (A-B) are stronger than the forces between molecules of the pure liquids (A-A and B-B).
- This stronger attraction makes it harder for molecules to escape into the vapor phase.
Effect on Vapor Pressure and Boiling Point
- Vapor Pressure : Since the molecules are held more tightly, the vapor pressure of the solution will be lower than expected from Raoult's law. Decreased vapor pressure.
- Boiling Point : A lower vapor pressure means you need to increase the temperature further to reach atmospheric pressure, where boiling occurs. Therefore, the boiling point of the solution will be higher than expected. Increased boiling point.
Answer
The correct answer is Option D: decreased vapor pressure, increased boiling point.
The vapor pressure of pure benzene and methyl benzene at $27^{\circ} \mathrm{C}$ is given as 80 Torr and 24 Torr, respectively. The mole fraction of methyl benzene in vapor phase, in equilibrium with an equimolar mixture of those two liquids (ideal solution) at the same temperature is _________ $\times 10^{-2}$ (nearest integer)
Explanation:
To find the mole fraction of methyl benzene in the vapor phase when it is in equilibrium with an equimolar (equal mole) mixture of benzene and methyl benzene at $27^{\circ}C$, we can apply Raoult's law for an ideal solution. Given the vapor pressures of pure benzene and methyl benzene are 80 Torr and 24 Torr, respectively.
Raoult's law states that the partial vapor pressure of a component in a solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution ($P_i = P_i^\circ x_i$), where $P_i$ is the partial vapor pressure of the component $i$, $P_i^\circ$ is the vapor pressure of the pure component $i$, and $x_i$ is the mole fraction of the component $i$ in the solution.
For an equimolar mixture of benzene and methyl benzene, the mole fractions of benzene ($x_b$) and methyl benzene ($x_{mb}$) in the liquid phase are both $0.5$, because the mixture is equimolar.
The total pressure of the mixture ($P_{total}$) can be calculated using Raoult's law for each component and summing up the individual pressures:
$P_{total} = P_{benzene} + P_{methyl\ benzene}$
$P_{total} = (P_b^\circ x_b) + (P_{mb}^\circ x_{mb})$
Substituting the given values:
$P_{total} = (80 \times 0.5) + (24 \times 0.5)$
$P_{total} = 40 + 12 = 52 \, \text{Torr}$
The mole fraction of methyl benzene in the vapor phase ($y_{mb}$) can be calculated using Dalton's law, which states that the partial pressure of a component in a mixture is equal to the mole fraction of that component in the vapor phase times the total pressure of the mixture.
The partial pressure of methyl benzene can be obtained from Raoult's law as we did earlier:
$P_{methyl\ benzene} = P_{mb}^\circ x_{mb} = 24 \times 0.5 = 12 \, \text{Torr}$
Using Dalton's law:
$y_{mb} = \frac{P_{methyl\ benzene}}{P_{total}}$
Substituting the values:
$y_{mb} = \frac{12}{52}$
$y_{mb} = \frac{6}{26}$
$y_{mb} = \frac{3}{13}$
To express $y_{mb}$ as a percentage times $10^{-2}$:
$y_{mb} = \left(\frac{3}{13}\right) \times 100 \times 10^{-2}$
$y_{mb} \approx 23.08 \times 10^{-2}$
So, the mole fraction of methyl benzene in the vapor phase, in equilibrium with an equimolar mixture of those two liquids at $27^{\circ}C$, is approximately $23 \times 10^{-2}$ (rounded to the nearest integer).
A solution containing $10 \mathrm{~g}$ of an electrolyte $\mathrm{AB}_2$ in $100 \mathrm{~g}$ of water boils at $100.52^{\circ} \mathrm{C}$. The degree of ionization of the electrolyte $(\alpha)$ is _________ $\times 10^{-1}$. (nearest integer)
[Given : Molar mass of $\mathrm{AB}_2=200 \mathrm{~g} \mathrm{~mol}^{-1}, \mathrm{~K}_{\mathrm{b}}$ (molal boiling point elevation const. of water) $=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$, boiling point of water $=100^{\circ} \mathrm{C} ; \mathrm{AB}_2$ ionises as $\mathrm{AB}_2 \rightarrow \mathrm{A}^{2+}+2 \mathrm{~B}^{-}]$
Explanation:
To find the degree of ionization $(\alpha)$, we need to use the boiling point elevation formula and the van't Hoff factor. The formulas we will use are:
$\Delta T_b = i \cdot K_b \cdot m$
where:
$\Delta T_b$ is the boiling point elevation,
$i$ is the van't Hoff factor,
$K_b$ is the molal boiling point elevation constant, and
$m$ is the molality of the solution.
Given data:
- $\Delta T_b = 100.52^{\circ} \mathrm{C} - 100^{\circ} \mathrm{C} = 0.52^{\circ} \mathrm{C}$
- Mass of $\mathrm{AB}_2$ = 10 g
- Mass of water = 100 g = 0.1 kg
- Molar mass of $\mathrm{AB}_2$ = 200 g/mol
- $K_b$ = 0.52 K kg mol-1
First, we calculate the molality (m):
$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}$
$= \frac{\frac{10}{200}}{0.1} \:\mathrm{mol\ kg}^{-1}$
$= 0.5 \:\mathrm{mol\ kg}^{-1}$
Next, using the formula for boiling point elevation:
$0.52 = i \cdot 0.52 \cdot 0.5$
$i = \frac{0.52}{0.52 \cdot 0.5}$
$i = \frac{1}{0.5}$
$i = 2$
Now, the van't Hoff factor $i$ is related to the degree of ionization $(\alpha)$. For the electrolyte $\mathrm{AB}_2$, which ionizes as $\mathrm{AB}_2 \rightarrow \mathrm{A}^{2+}+2\mathrm{B}^{-}$, the van't Hoff factor $i$ can be expressed as:
$i = 1 + (n-1)\alpha$
where
- $n$ is the number of ions produced (which is 3 for $\mathrm{AB}_2 \rightarrow \mathrm{A}^{2+}+2\mathrm{B}^{-}$)
- $\alpha$ is the degree of ionization
Substituting $i = 2$ into the equation:
$2 = 1 + 2 \alpha$
$2 - 1 = 2 \alpha$
$\alpha = \frac{1}{2}$
$\alpha = 0.5$
The degree of ionization $(\alpha)$ is therefore
$0.5 \times 10^{-1} = 5 \times 10^{-2}$.
Hence, the degree of ionization of the electrolyte $(\alpha)$ is approximately 5 $\times 10^{-1}$ (the nearest integer is 5).
When '$x$' $\times 10^{-2} \mathrm{~mL}$ methanol (molar mass $=32 \mathrm{~g}$' density $=0.792 \mathrm{~g} / \mathrm{cm}^3$) is added to $100 \mathrm{~mL}$. water (density $=1 \mathrm{~g} / \mathrm{cm}^3$), the following diagram is obtained.
$x=$ ________ (nearest integer).
[Given : Molal freezing point depression constant of water at $273.15 \mathrm{~K}$ is $1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$]
Explanation:
$\begin{aligned} & \Delta T_f=2.5^{\circ} \mathrm{C} \\ & \Delta T_f=i \times k_f \times m \\ & 2.5=1 \times 1.86 \times \frac{n_B}{0.1} \\ & n_B=\frac{2.5 \times 0.1}{1.86}=0.1344 \mathrm{~mol} \\ & \text { mass of methanol }=0.1344 \times 32=4.3 \mathrm{~g} \\ & d=\frac{m}{v} \\ & v=\frac{m}{d} \\ & v=\frac{4.3}{0.792} \mathrm{~mL} \\ & =5.43 \mathrm{~mL} \\ & =543 \times 10^{-2} \mathrm{~mL} \\ & x=543 \end{aligned}$
Considering acetic acid dissociates in water, its dissociation constant is $6.25 \times 10^{-5}$. If $5 \mathrm{~mL}$ of acetic acid is dissolved in 1 litre water, the solution will freeze at $-x \times 10^{-2}{ }^{\circ} \mathrm{C}$, provided pure water freezes at $0{ }^{\circ} \mathrm{C}$.
$x=$ _________. (Nearest integer)
$\begin{aligned} \text{Given :} \quad & \left(\mathrm{K}_{\mathrm{f}}\right)_{\text {water }}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}-1 \\ & \text { density of acetic acid is } 1.2 \mathrm{~g} \mathrm{~mol}^{-1} \text {. } \\ & \text { molar mass of water }=18 \mathrm{~g} \mathrm{~mol}^{-1} \text {. } \\ & \text { molar mass of acetic acid= } 60 \mathrm{~g} \mathrm{~mol}^{-1} \text {. } \\ & \text { density of water }=1 \mathrm{~g} \mathrm{~cm}^{-3} \end{aligned}$
Acetic acid dissociates as $\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{\ominus}+\mathrm{H}^{\oplus}$
Explanation:
To solve the problem, we'll calculate the freezing point depression of the acetic acid solution in water.
- Calculating Moles of Acetic Acid:
- Given volume of acetic acid: $ 5 \, \text{mL} $
- Density of acetic acid: $ 1.2 \, \text{g/mL} $
- Molar mass of acetic acid: $ 60 \, \text{g/mol} $
$ \text{Mass of acetic acid} = \text{Volume} \times \text{Density} = 5 \, \text{mL} \times 1.2 \, \text{g/mL} = 6 \, \text{g} $
$ \text{Moles of acetic acid} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{6 \, \text{g}}{60 \, \text{g/mol}} = 0.1 \, \text{mol} $
- Calculating Molality:
- Volume of water: $ 1 \, \text{L} $
- Density of water: $ 1 \, \text{g/cm}^3 \Rightarrow 1 \, \text{kg/L} $
- Hence, mass of water = $ 1 \, \text{kg} $
$ \text{Molality (m)} = \frac{\text{Moles of solute}}{\text{Mass of solvent (kg)}} = \frac{0.1 \, \text{mol}}{1 \, \text{kg}} = 0.1 \, \text{mol/kg} $
- Degree of Dissociation ($\alpha$):
- Dissociation constant ($K_a$) of acetic acid: $6.25 \times 10^{-5}$
$ \alpha = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{6.25 \times 10^{-5}}{0.1}} = \sqrt{6.25 \times 10^{-4}} = 25 \times 10^{-3} = 0.025 $
- Van't Hoff factor (i):
- Acetic acid partially dissociates into 2 ions (CH$_3$COO$^-$ and H$^+$)
$ i = 1 + (\text{number of ions produced} - 1) \times \alpha = 1 + (2-1) \times 0.025 = 1 + 0.025 = 1.025 $
- Freezing Point Depression ($\Delta T_f$):
- Freezing point depression constant ($K_f$) for water: $1.86 \,\text{K kg/mol}$
$ \Delta T_f = i \times K_f \times \text{Molality} = 1.025 \times 1.86 \times 0.1 = 0.19065 \, \text{K} $
- Final Freezing Point:
- Pure water freezes at $0 \, ^\circ \text{C}$
$ \text{Freezing point of solution} = 0 \, ^\circ \text{C} - 0.19065 \, ^\circ \text{C} = -0.19065 \, ^\circ \text{C} $
Here, $-x \times 10^{-2} \, ^\circ \text{C}$ is given. Therefore, $x = 19$ (nearest integer).
So, the final answer is:
$ x = 19 $