Solutions

$ \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

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)

$ \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} $

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)

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) $2.46\ \text{atm}$

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

Mixture of $\mathrm{CS}_2$ and

show positive deviation $\mathrm{P}_{\mathrm{CS}_2}>\mathrm{P}_{\mathrm{CS}_2}^{\mathrm{o}} \cdot \mathrm{X}_{\mathrm{CS}_2}$

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 $\mathrm{H}_2 \mathrm{SO}_4=\frac{\frac{147}{98}}{\left(\frac{147}{98}+\frac{53}{18}\right)}=0.337$

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 $\log K_H$.

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).

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)

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} $

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 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.

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)

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

Normal osmosis occurs from (2) to (1)

For reverse osmosis from (1) to (2)

Pressure : $\mathrm{P}_1>\pi$

$\therefore$ Answer [A & C] only

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 $377.3 \, \text{K}$.

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.

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.

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.

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 $\mathrm{He}>\mathrm{N}_2>$ $\mathrm{CH}_4$.

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.

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.

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

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} $.

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 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.

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 $\frac{\mathrm{P}_2^{\mathrm{o}} - \mathrm{P}_1^{\mathrm{o}}}{\mathrm{P}_2^{\mathrm{o}}}$.

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 $i = 1.$

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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}. $

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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 $(\text{ii}) < (\text{i}) < (\text{iii}) < (\text{iv})$.

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Final Answer

$ \boxed{\text{(ii) } < \text{(i) } < \text{(iii) } < \text{(iv)}} \quad \text{(Option B)} $

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.

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.

$\left(\mathrm{CH}_3\right)_2 \mathrm{CO}+\mathrm{CS}_2$ Exibits positive deviations from Raoult's Law

$\Delta \mathrm{T}_{\mathrm{f}}$ 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, $\mathrm{k}_{\mathrm{f}}{ }^{\prime}\left(\mathrm{H}_2 \mathrm{O}\right)=1.86 \mathrm{~k} \cdot \mathrm{~kg} \mathrm{~mol}^{-1}$

$\mathrm{k}_{\mathrm{f}}{ }^{\prime}(\text { Benzene })=5.12 \mathrm{~k} \cdot \mathrm{~kg} \mathrm{~mol}^{-1}$

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

Negative Deviation from Raoult's Law

• 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 lowering of vapor pressure of a solvent by a nonvolatile solute is given by Raoult's law:

$\Delta P = x_{\text{solute}} \cdot P_0$

where $\Delta P$ is the change in vapor pressure, $x_{\text{solute}}$ is the mole fraction of the solute, and $P_0$ is the vapor pressure of the pure solvent.

Rearranging the formula for $x_{\text{solute}}$, we have:

$x_{\text{solute}} = \frac{\Delta P}{P_0}$

Substituting the given values, we get :

$x_{\text{solute}} = \frac{0.20 \, \text{mm Hg}}{54.2 \, \text{mm Hg}} = 0.003689$

Since the mole fraction of the solute is also equal to the number of moles of solute divided by the total number of moles, we can express $x_{\text{solute}}$ as:

$x_{\text{solute}} = \frac{\text{moles}_{\text{solute}}}{\text{moles}_{\text{solute}} + \text{moles}_{\text{water}}}$

Assuming that the solution is dilute, the number of moles of water will be much larger than the number of moles of solute, so we can approximate the total number of moles as the number of moles of water.

Thus, we have :

$x_{\text{solute}} \approx \frac{\text{moles}_{\text{solute}}}{\text{moles}_{\text{water}}}$

Therefore, the number of moles of solute is :

$\text{moles}_{\text{solute}} = x_{\text{solute}} \cdot \text{moles}_{\text{water}}$

The number of moles of water is the mass of the water divided by the molar mass of water (18 g/mol) :

$\text{moles}_{\text{water}} = \frac{100 \, \text{g}}{18 \, \text{g/mol}} = 5.56 \, \text{mol}$

So, the number of moles of solute is :

$\text{moles}_{\text{solute}} = 0.003689 \cdot 5.56 \, \text{mol} = 0.0205 \, \text{mol}$

Finally, to find the mass of the glucose, we multiply the number of moles of glucose by the molar mass of glucose :

$\text{mass}_{\text{glucose}} = \text{moles}_{\text{solute}} \cdot \text{molar mass}_{\text{glucose}}$

$\text{mass}_{\text{glucose}} = 0.0205 \, \text{mol} \cdot 180 \, \text{g/mol} = 3.69 \, \text{g}$

So, the correct answer is 3.69 g.

(A) Elevation in boiling point temperature of water will be higher for 0.1 M NaCl as compared to 0.1 M urea.

(B) Azeotropic mixtures boil without change in their composition

(C) Osmosis always takes place from hypotonic (low concentration of solute) solution to hypertonic (high concentration of solute) solution.

(E) When KI solution is added to AgNO₃ solution, positively charged solution is formed due to adsorption of Ag⁺ ions from dispersion medium.

AgI/ Ag⁺ (positively charged)

(D) Let the mass of H₂SO₄ (32%) is 100.

$ \therefore \text { Weight of } \mathrm{H}_2 \mathrm{SO}_4=32 $

Moles of $\mathrm{H}_2 \mathrm{SO}_4=\frac{32}{98}$

Now, $4.09=\frac{32}{98 \times \mathrm{V}} \Rightarrow \mathrm{V}=79 \mathrm{ml}$

Density $=\frac{100}{79}=1.265$

Hence, correct answer is (C) B and D only.

A.	van't Hoff factor	III.	$\mathrm{{{Normal\,molar\,mass} \over {Abnormal\,molar\,mass}}}$
B.	$\mathrm{k_f}$	I.	Cryoscopic constant
C.	Solutions with same osmotic pressure	II.	Isotonic solutions
D.	Azeotropes	IV.	Solutions with same composition of vapour above it

In the depression of freezing point experiment only solvent molecules solidify and vapour pressure of solution decreases as some of the surface area is occupied by solute molecules, so less number of molecules will go in vapour form.

For $\mathbf{A}: 100 \,\mathrm{gm}$ solution $\rightarrow 2 \,\mathrm{gm}$ solute $\mathrm{A}$

$\therefore$ Molality $=\frac{2 / \mathrm{M}_{\mathrm{A}}}{0.098}$

For B : $100 \,\mathrm{gm}$ solution $\rightarrow 8 \,\mathrm{gm}$ solute $\mathrm{B}$

$ \begin{aligned} &\therefore \text { Molality }=\frac{8 / \mathrm{M}_{\mathrm{B}}}{0.092} \\\\ &\because\left(\Delta \mathrm{T}_{\mathrm{B}}\right)_{\mathrm{A}}=\left(\Delta \mathrm{T}_{\mathrm{B}}\right)_{\mathrm{B}} \end{aligned} $

$\therefore$ Molality of $\mathrm{A}=$ Molality of $\mathrm{B}$

$ \therefore \frac{2}{0.098 \mathrm{M}_{\mathrm{A}}}=\frac{8}{0.092 \mathrm{M}_{\mathrm{B}}} $

$ \frac{2}{98} \times \frac{92}{8}=\frac{M_A}{M_B} $

$ \frac{1}{4.261}=\frac{\mathrm{M}_{\mathrm{A}}}{\mathrm{M}_{\mathrm{B}}} $

$ \therefore \mathrm{M}_{\mathrm{B}}=4.261 \times \mathrm{M}_{\mathrm{A}} $

$\Delta T_{f}=i k_{f} \times m$

$ \frac{\Delta T_{\mathrm{f}(\mathrm{A})}}{\Delta \mathrm{T}_{\mathrm{f}(\mathrm{B})}}=\frac{1}{4} $

$\frac{\mathrm{i} \times \mathrm{K}_{\mathrm{f}} \times \frac{1}{\mathrm{M}_{\mathrm{A}}} \times 1}{\mathrm{i} \times \mathrm{K}_{\mathrm{f}} \times \frac{1}{\mathrm{M}_{\mathrm{B}}} \times 1}=\frac{1}{4}$

$\frac{M_{B}}{M_{A}}=\frac{1}{4}$

$M_{A}: M_{B}=4: 1$

$\Delta \mathrm{T}_{\mathrm{f}}$ of formic acid $=0.0405^{\circ} \mathrm{C}$

Concentration $=0.5 \mathrm{~mL} / \mathrm{L}$

and density $=1.05 \mathrm{~g} / \mathrm{mL}$

$\therefore$ Mass of formic acid in solution $=1.05 \times 0.5 \mathrm{~g}$

$ =0.525 \mathrm{~g} $

$\therefore$ According to Van't Hoff equation,

$ \begin{aligned} &\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \mathrm{k}_{\mathrm{f}} \cdot \mathrm{m} \\ &0.0405=\mathrm{i} \times 1.86 \times \frac{0.525}{46 \times 1} \end{aligned} $

(Assuming mass of $1 \mathrm{~L}$ water $=\mathrm{kg}$ )

$ \mathrm{i}=\frac{0.0405 \times 46}{1.86 \times 0.525}=1.89 \approx 1.9 $

According to Henry's law $p={K_{H}} \times x$

for $x$ constant $: p ~ \alpha ~ K_{\mathrm{i}}$

$p$ has max partial pressures.

So it should have maximum $K_{\mathrm{H}}$ and $s$ have minimum partial pressure, So $K_{H}$ should be minimum.

Molality and Mass are temperature Independent so on changing temp., molality and mass remain unchanged.

Since, $\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{ik}_{\mathrm{f} m}$

$ \begin{aligned} &m=\frac{0.7}{93} \times \frac{1000}{42} \\\\ &0.2=i \times 1.86 \times \frac{0.7 \times 1000}{93 \times 42} \\\\ &i=0.6 \\\\ &\alpha=\frac{i-1}{\frac{1}{n}-1}=\frac{0.6-1}{\frac{1}{2}-1}=0.8 \end{aligned} $

Hence, the percentage association of solute $A$ is $80 \%$.

ΔT_f $ \propto $ i × m

greater the value of i, greater will be the ΔT_f value.

Van't Hoff factor is highest for KHSO₄

$\therefore$ colligative property ($\Delta$T_f) will be highest for KHSO₄

To determine which solution has the lowest freezing point, you should consider the van 't Hoff factor $i$, which represents the number of particles the solute dissociates into when dissolved. The freezing point depression is given by:

$ \Delta T_f = i \cdot K_f \cdot m $

where:

$\Delta T_f$ is the change in freezing point,

$K_f$ is the freezing point depression constant,

$m$ is the molality of the solution,

$i$ is the van 't Hoff factor, which depends on the dissociation of the solute.

For each solute, the van 't Hoff factor $i$ can be determined as follows:

Option A: $\text{Al}_2(\text{SO}_4)_3$

Dissociates into 2 Al$^{3+}$ ions and 3 SO$_4^{2-}$ ions.

$i = 2 + 3 = 5$

Option B: $\text{C}_6\text{H}_{12}\text{O}_6$ (glucose)

Does not dissociate in solution (non-electrolyte).

$i = 1$

Option C: $\text{KI}$

Dissociates into K$^+$ and I$^-$ ions.

$i = 1 + 1 = 2$

Option D: $\text{K}_2\text{SO}_4$

Dissociates into 2 K$^+$ ions and 1 SO$_4^{2-}$ ion.

$i = 2 + 1 = 3$

Comparing the van 't Hoff factors, the solution of $\text{Al}_2(\text{SO}_4)_3$ will have the highest factor ($i = 5$), leading to the greatest freezing point depression. Thus, the $\text{Al}_2(\text{SO}_4)_3$ solution will have the lowest freezing point since the extent of freezing point depression is directly proportional to $i$.

Therefore, the solution of $\text{Al}_2(\text{SO}_4)_3$ has the lowest freezing point.

Relative lowering in vapour pressure (RLVP)

= ${{P - {P_s}} \over P} = {n \over {n + N}}$

n $ \to $ moles of solute
N $ \to $ moles of solvent

$ \therefore $ (RLVP)_A = ${{{{10} \over {100}}} \over {{{10} \over {100}} + {{180} \over {18}}}}$

(RLVP)_B = ${{{{10} \over {200}}} \over {{{10} \over {200}} + {{180} \over {18}}}}$

and (RLVP)_C = ${{{{10} \over {10000}}} \over {{{10} \over {10000}} + {{180} \over {18}}}}$

$ \therefore $ (RLVP)_A > (RLVP)_B > (RLVP)_C

So, A > B > C

Raw mango shrink in salt solution due to net transfer of water molecules from mango to salt solution due to phenomenon of osmosis.