Mix Examples of Solutions Questions in English

(B) Given $pH = 2$.
We know that $pH = -\log [H^+]$,so $[H^+] = 10^{-pH} = 10^{-2} = 0.01 \ M$.
For a monobasic acid $HX$,the dissociation is $HX \rightleftharpoons H^+ + X^-$.
The degree of dissociation $\alpha$ is given by $[H^+] = c \alpha$,where $c = 0.1 \ M$.
$\alpha = \frac{[H^+]}{c} = \frac{0.01}{0.1} = 0.1$.
The van't Hoff factor $i$ for a binary electrolyte is $i = 1 + \alpha = 1 + 0.1 = 1.1$.
The osmotic pressure $\Pi$ is given by $\Pi = iCRT$.
Substituting the values: $\Pi = 1.1 \times 0.1 \times RT = 0.11 \ RT$.
Thus,the correct option is $B$.

(C) The depression in freezing point is given by $\Delta T_f = T_f^\circ - T_f = 0 - (-0.186) = 0.186 \ ^\circ C$.
We know that $\Delta T_f = K_f \times m$ and $\Delta T_b = K_b \times m$.
Dividing the two equations: $\frac{\Delta T_b}{\Delta T_f} = \frac{K_b}{K_f}$.
Substituting the values: $\Delta T_b = \Delta T_f \times \frac{K_b}{K_f} = 0.186 \times \frac{0.512}{1.86}$.
$\Delta T_b = 0.1 \times 0.512 = 0.0512 \ ^\circ C$.

(B) The osmotic pressure is given by the formula $\pi = iCRT$.
Given $pH = 2$,the concentration of hydrogen ions is $[H^{+}] = 10^{-2} \, M$.
For a monobasic acid $HA \rightleftharpoons H^{+} + A^{-}$,the degree of dissociation $\alpha$ is calculated as $\alpha = \frac{[H^{+}]}{C} = \frac{10^{-2}}{0.1} = 0.1$.
The van't Hoff factor $i$ for dissociation is $i = 1 + \alpha(n - 1)$,where $n = 2$.
Thus,$i = 1 + 0.1(2 - 1) = 1.1$.
Substituting the values into the osmotic pressure formula: $\pi = 1.1 \times 0.1 \times RT = 0.11 \, RT$.

(A) The elevation in boiling point is given by $\Delta T_b = i K_b m$.
Given $\Delta T_b = 0.75 \, ^oC$,$K_b = 5 \, K \, kg \, mol^{-1}$,and $m = 0.1 \, m$.
Substituting these values: $0.75 = i \times 5 \times 0.1$.
$i = \frac{0.75}{0.5} = 1.5$.
For the dissociation of $CH_3COOH \rightleftharpoons CH_3COO^- + H^+$,the van't Hoff factor $i = 1 + \alpha$.
$1.5 = 1 + \alpha \implies \alpha = 0.5$.
The dissociation constant $K_a$ is given by $K_a = \frac{c \alpha^2}{1 - \alpha}$.
$K_a = \frac{0.1 \times (0.5)^2}{1 - 0.5} = \frac{0.1 \times 0.25}{0.5} = 0.05 = 5 \times 10^{-2}$.

(D) As vapours are removed,the remaining solution becomes more concentrated.
The freezing point depression is defined as $\Delta T_f = K_f \times m$. As the molality $m$ increases,$\Delta T_f$ increases,which means the freezing point $(T_f = T_f^0 - \Delta T_f)$ decreases.
Osmotic pressure is given by $\Pi = CRT$. As the concentration $C$ increases,the osmotic pressure $\Pi$ increases.
Therefore,the freezing point decreases and the osmotic pressure increases.

(A) The elevation in boiling point is given by $\Delta T_b = i K_b m$.
For $Ag_2SO_4$,the dissociation is $Ag_2SO_4 \rightleftharpoons 2Ag^+ + SO_4^{2-}$,so the van't Hoff factor $i = 3$.
Given $\Delta T_b = 0.003 \ K$,$K_b = 5 \ K \ kg \ mol^{-1}$,and $m = S$ (solubility).
Substituting the values: $0.003 = 3 \times 5 \times S$.
$S = \frac{0.003}{15} = 2 \times 10^{-4} \ mol \ L^{-1}$.
The solubility product $K_{sp}$ for $Ag_2SO_4$ is $K_{sp} = [Ag^+]^2 [SO_4^{2-}] = (2S)^2 (S) = 4S^3$.
$K_{sp} = 4 \times (2 \times 10^{-4})^3 = 4 \times 8 \times 10^{-12} = 3.2 \times 10^{-11}$.

(A) For freezing point: Theoretical $\Delta T_f = K_f \times m = 5.12 \times 1 = 5.12 \ ^{\circ}C$.
Observed $\Delta T_f = 2.56 \ ^{\circ}C$.
Since observed $\Delta T_f < $ theoretical $\Delta T_f$,the van't Hoff factor $i = \frac{2.56}{5.12} = 0.5$,which indicates dimer formation.
For boiling point: Theoretical $\Delta T_b = K_b \times m = 2.53 \times 1 = 2.53 \ ^{\circ}C$.
Observed $\Delta T_b = 2.53 \ ^{\circ}C$.
Since observed $\Delta T_b =$ theoretical $\Delta T_b$,$i = 1$,indicating no association or dissociation.
Thus,Statement $I$ and Statement $II$ are correct.

(C) For the formation of an ideal binary solution:
$1$. The enthalpy of mixing is zero,i.e.,$\Delta H_{mix} = 0$.
$2$. The entropy of mixing is positive,i.e.,$\Delta S_{mix} > 0$,so $T \Delta S_{mix} > 0$.
$3$. The Gibbs free energy of mixing is given by $\Delta G_{mix} = \Delta H_{mix} - T \Delta S_{mix}$. Since $\Delta H_{mix} = 0$,we have $\Delta G_{mix} = -T \Delta S_{mix}$. Because $\Delta S_{mix} > 0$,$\Delta G_{mix}$ must be negative $(\Delta G_{mix} < 0)$.
Comparing these conditions with the given options:
- $\Delta H_{mix}$ should be a straight line at $0$.
- $T \Delta S_{mix}$ should be a positive curve (above the $0$ axis).
- $\Delta G_{mix}$ should be a negative curve (below the $0$ axis).
Option $C$ correctly shows $\Delta H_{mix} = 0$,$T \Delta S_{mix} > 0$,and $\Delta G_{mix} < 0$.

(C) According to Raoult's law,the partial pressure of $A$ is given by $P_A = P^o_A \cdot x_A$,where $x_A$ is the mole fraction of $A$ in the liquid phase.
Since $x$ is the mole fraction of $B$ in the liquid phase,$x_A = 1 - x$.
Thus,$P_A = P^o_A(1 - x) = P^o_A - P^o_A \cdot x$.
Also,from Dalton's law,$P_A = y \cdot P_{total}$,where $y$ is the mole fraction of $A$ in the vapour phase.
Given the equation $P_A = 235y - 125xy$,we can rewrite this as $P_A = y(235 - 125x)$.
Comparing the expressions,we note that for a binary mixture,$P_{total} = P^o_A x_A + P^o_B x_B = P^o_A(1-x) + P^o_B x = P^o_A + (P^o_B - P^o_A)x$.
Substituting this into $P_A = y \cdot P_{total}$,we get $P^o_A(1-x) = y(P^o_A + (P^o_B - P^o_A)x)$.
Rearranging to match the form $P_A = 235y - 125xy$,we identify $P^o_A = 235$ and $P^o_A - P^o_B = 125$.
Therefore,$P^o_B = P^o_A - 125 = 235 - 125 = 110 \, \text{mm of Hg}$.

(B) For boiling point elevation: $\Delta T_b = i_b \times m \times K_b$
$101.08 - 100 = 2 \times m \times K_b$ (Since $AB$ is $100\%$ ionized at $b.p.$,$i_b = 2$)
$1.08 = 2 \times m \times K_b \implies m \times K_b = 0.54$
For freezing point depression: $\Delta T_f = i_f \times m \times K_f$
$0 - (-1.80) = i_f \times m \times (K_b / 0.3)$ (Given $K_b / K_f = 0.3 \implies K_f = K_b / 0.3$)
$1.80 = i_f \times (0.54 / 0.3)$
$1.80 = i_f \times 1.8 \implies i_f = 1$
Since $i_f = 1$,$AB$ behaves as a non-electrolyte at the freezing point.

(A) For salt $MX_4$,the dissociation is $MX_4 \rightarrow M^{4+} + 4X^-$. The number of ions $n = 5$.
Van't Hoff factor $i = 1 + (n-1)\alpha = 1 + (5-1)(0.9) = 1 + 3.6 = 4.6$.
Freezing point depression $\Delta T_f = T_f^0 - T_f = 0 - (-3) = 3\ ^oC$.
Using $\Delta T_f = i \cdot k_f \cdot m$,we get $3 = 4.6 \cdot 1.86 \cdot m$,so $m = 3 / (4.6 \cdot 1.86) \approx 0.3506\ mol\ kg^{-1}$.
At boiling point,the salt is $100\%$ ionized,so $i' = n = 5$.
Boiling point elevation $\Delta T_b = i' \cdot k_b \cdot m = 5 \cdot 0.52 \cdot 0.3506 \approx 0.911\ ^oC$.
Boiling point of solution $= 100 + 0.911 = 100.911\ ^oC \approx 100.9\ ^oC$.

(A) Given: $m = 0.025 \ m$,$\Delta T_f = 0 - (-0.060) = 0.060 \ K$,$K_f = 1.86 \ K \ kg \ mol^{-1}$.
Using the formula $\Delta T_f = i \times K_f \times m$:
$0.060 = i \times 1.86 \times 0.025$
$i = \frac{0.060}{0.0465} \approx 1.29$.
For a monobasic acid $HA \rightleftharpoons H^+ + A^-$,the van't Hoff factor $i = 1 + \alpha$,where $\alpha$ is the degree of dissociation.
$1 + \alpha = 1.29 \implies \alpha = 0.29$.
The ionization constant $K_a = \frac{c \alpha^2}{1 - \alpha} = \frac{0.025 \times (0.29)^2}{1 - 0.29} = \frac{0.025 \times 0.0841}{0.71} \approx 0.00296 \approx 3 \times 10^{-3}$.

(C) Let us analyze each option:
$A$: Acidic strength depends on bond dissociation energy. $HCl$ has a stronger bond than $HBr$,so $HBr$ is more acidic. Reducing property also follows the order $HI > HBr > HCl > HF$. Thus,both properties follow the same order. This is incorrect.
$B$: Basic strength depends on the availability of the lone pair. As we go down the group,the lone pair is in a larger orbital,making it less available. Thus,$PH_3 > AsH_3$. Bond angle also follows the same order due to the decrease in electronegativity of the central atom. This is incorrect.
$C$: The dipole moment of $CH_3Cl$ $(1.86 \ D)$ is greater than $CH_3F$ $(1.85 \ D)$ due to the larger bond length in $C-Cl$. The $H-C-H$ bond angle in $CH_3F$ is larger than in $CH_3Cl$ due to the higher electronegativity of $F$ causing more $s$-character in $C-H$ bonds. This is correct.
$D$: Maleic acid (cis-isomer) has a higher $Ka_1$ due to intramolecular hydrogen bonding in the mono-anion,but its $Ka_2$ is lower than fumaric acid due to strong electrostatic repulsion between the two carboxylate groups. This is correct.
Note: Both $C$ and $D$ are scientifically correct statements.

(D) Given: $\Delta T_b = 0.78 \, K$,$K_b = 0.52 \, K \, kg \, mol^{-1}$,$K_f = 1.86 \, K \, kg \, mol^{-1}$.
We know that $\Delta T_b = K_b \times m$ and $\Delta T_f = K_f \times m$.
Dividing the two equations: $\frac{\Delta T_f}{\Delta T_b} = \frac{K_f}{K_b}$.
$\Delta T_f = \Delta T_b \times \frac{K_f}{K_b} = 0.78 \times \frac{1.86}{0.52} = 0.78 \times 3.5769 \approx 2.79 \, K$.
The freezing point of pure water is $273.15 \, K$.
The freezing point of the solution = $T_f^0 - \Delta T_f = 273.15 - 2.79 = 270.36 \, K$.
Rounding to the nearest provided option,the correct value is $270.21 \, K$.

(B) Step $1$: Calculate the molality $(m)$ of the solution using the freezing point depression formula: $\Delta T_f = K_f \times m$. Given $\Delta T_f = 0.52\,^oC$ and $K_f = 1.86\,^oC\, m^{-1}$,we have $0.52 = 1.86 \times m$,so $m = \frac{0.52}{1.86} \approx 0.2796 \, m$.
Step $2$: Since molality and molarity are assumed to be numerically equal,$M \approx 0.2796 \, M$.
Step $3$: Calculate the osmotic pressure $(\pi)$ using the formula $\pi = MRT$,where $R = 0.0821 \, L \, atm \, K^{-1} \, mol^{-1}$ and $T = 37 + 273 = 310 \, K$.
Step $4$: $\pi = 0.2796 \times 0.0821 \times 310 \approx 7.11 \, atm$.

(A) Let the mass of urea be $w \ g$. The molar mass of urea $(NH_2CONH_2)$ is $60 \ g \ mol^{-1}$.
Molality $(m) = \frac{w \times 1000}{60 \times 10000} = \frac{w}{600} \ mol \ kg^{-1}$.
Elevation in boiling point: $\Delta T_b = K_b \times m = 0.513 \times \frac{w}{600}$.
Depression in freezing point: $\Delta T_f = K_f \times m = 1.86 \times \frac{w}{600}$.
The boiling point of the solution is $T_b = 100 + \Delta T_b$ and the freezing point is $T_f = 0 - \Delta T_f$.
The difference is $(100 + \Delta T_b) - (0 - \Delta T_f) = 100 + \Delta T_b + \Delta T_f = 100.2372$.
Therefore,$\Delta T_b + \Delta T_f = 0.2372$.
$\frac{w}{600} \times (0.513 + 1.86) = 0.2372$.
$\frac{w}{600} \times 2.373 = 0.2372$.
$w = \frac{0.2372 \times 600}{2.373} \approx 60 \ g$.
Comparing with options,the closest value is $59.64 \ g$.

(C) Proof spirit is defined as a mixture of ethyl alcohol and water that contains $57.1\%$ of ethyl alcohol by volume.
This corresponds to approximately $49.28\%$ by mass.
Therefore,the correct composition is $57.1\% \text{ ethyl alcohol}$ and $42.9\% \text{ water}$.

(D) For $NaCl$,the van't Hoff factor $i_1 = 2$. The depression in freezing point is given by $\Delta T_f = i_1 K_f m = 0.372 \, K$.
Thus,$m = \frac{0.372}{2 \times 1.86} = 0.1 \, mol \, kg^{-1}$.
For $BaCl_2$,the van't Hoff factor $i_2 = 3$.
The elevation in boiling point is $\Delta T_b = i_2 K_b m = 3 \times 0.52 \times 0.1 = 0.156 \, K$.
The boiling point of the solution is $T_b = 100 + \Delta T_b = 100 + 0.156 = 100.156 \, ^oC$.

(B) Given: Boiling point of solution = $100.18 \, ^oC$. Boiling point of pure water = $100 \, ^oC$.
Elevation in boiling point,$\Delta T_b = 100.18 - 100 = 0.18 \, ^oC$.
We know that $\Delta T_b = K_b \cdot m$ and $\Delta T_f = K_f \cdot m$.
Therefore,$\frac{\Delta T_f}{\Delta T_b} = \frac{K_f}{K_b}$.
$\Delta T_f = \Delta T_b \times \frac{K_f}{K_b} = 0.18 \times \frac{1.86}{0.512} \approx 0.654 \, ^oC$.
Since $\Delta T_f = T_f^{\circ} - T_f$,where $T_f^{\circ} = 0 \, ^oC$ for water:
$0.654 = 0 - T_f$.
$T_f = -0.654 \, ^oC$.
Rounding to two decimal places,the freezing point is $-0.65 \, ^oC$.

(D) The depression in freezing point is given by $\Delta T_f = T_f^o - T_f = 0 - (-0.465) = 0.465 \ K$.
Using the formula for molar mass $M$ of the solute:
$M = \frac{1000 \times K_f \times w_2}{\Delta T_f \times w_1} = \frac{1000 \times 1.86 \times 1.8}{0.465 \times 40} = 180 \ g \ mol^{-1}$.
The empirical formula mass of $CH_2O$ is $12 + (2 \times 1) + 16 = 30 \ g \ mol^{-1}$.
Calculating the value of $n$:
$n = \frac{\text{Molecular mass}}{\text{Empirical formula mass}} = \frac{180}{30} = 6$.
Therefore,the molecular formula is $(CH_2O)_6 = C_6H_{12}O_6$.

(B) Both assertion and reason are correct but reason is not the correct explanation of assertion.
The relationship between lowering of vapour pressure and osmotic pressure can be derived as follows:
Van't Hoff equation for a dilute solution is $\pi = \frac{n}{V}RT$ ..... $(i)$
In the case of a dilute solution,the volume of the solution can be taken as equal to that of the solvent. If $N$ is the number of moles of solvent of molecular weight $M$ and density $\rho$,the volume $V$ is given by $V = \frac{NM}{\rho}$ ...... $(ii)$
Substituting $(ii)$ in $(i)$,we get $\frac{n}{N} = \frac{\pi M}{\rho RT}$ .... $(iii)$
From Raoult's law,the relative lowering of vapour pressure is $\frac{p^o - p}{p^o} = \frac{n}{N}$ ...... $(iv)$
Equating $(iii)$ and $(iv)$,we get $\frac{p^o - p}{p^o} = \frac{\pi M}{\rho RT}$
Rearranging,$(p^o - p) = \frac{\pi M p^o}{\rho RT}$
Since $\frac{M p^o}{\rho RT}$ is constant at a constant temperature,$(p^o - p) \propto \pi$.
Thus,the lowering of vapour pressure is directly proportional to the osmotic pressure. The assertion is correct.
Osmotic pressure is indeed a colligative property,so the reason is also correct. However,the fact that osmotic pressure is a colligative property does not explain why the lowering of vapour pressure is proportional to it.

(D) According to the thermodynamic relationship between Raoult's law and Henry's law,if one component of a binary solution obeys Raoult's law $(P_i = x_i P_i^o)$ over the entire range of composition,the other component must also obey Raoult's law.
However,in dilute solutions,the solvent obeys Raoult's law $(P_1 = x_1 P_1^o)$ while the solute obeys Henry's law $(P_2 = K_H x_2)$.
The Assertion is incorrect because if one component obeys Raoult's law,the other component often obeys Henry's law in the dilute range.
The Reason is correct because Raoult's law is indeed a special case of Henry's law where the Henry's constant $(K_H)$ becomes equal to the pure component vapor pressure $(P_i^o)$.

(C) The observed total vapour pressure $(600 \; mm \; Hg)$ is greater than the vapour pressure of either pure component ($512 \; mm \; Hg$ and $344 \; mm \; Hg$).
This indicates that the solution exhibits a positive deviation from Raoult's law.
For solutions showing positive deviation:
$1$. $\Delta_{sol} H > 0$ (Endothermic process,heat is absorbed).
$2$. $\Delta_{sol} V > 0$ (Volume increases,so the final volume is $> 200 \; mL$).
$3$. The solute-solvent interactions are weaker than the solute-solute and solvent-solvent interactions.
$4$. Raoult's law is not obeyed.
Therefore,the statement that the volume is $< 200 \; mL$ is false.

(C) $1$. Calculate the freezing point of the solution:
$\Delta T_{f} = K_{f} \times m = 2.0 \; K \; kg \; mol^{-1} \times 0.5 \; mol \; kg^{-1} = 1.0 \; K$.
Freezing point of solution $T = 273 - 1 = 272 \; K$.
$2$. Calculate the initial pressure of the gas:
Using $PV = nRT$,$P = \frac{nRT}{V} = \frac{0.1 \; mol \times 0.08 \; dm^{3} \; atm \; K^{-1} \; mol^{-1} \times 272 \; K}{1.0 \; dm^{3}} = 2.176 \; atm$.
$3$. After withdrawing the stoppers,the piston is frictionless and exposed to atmospheric pressure $(P_{ext} = 1 \; atm)$. The gas will expand until its internal pressure equals the external pressure.
Using Boyle's Law $(P_{1}V_{1} = P_{2}V_{2})$:
$2.176 \; atm \times 1.0 \; L = 1.0 \; atm \times V_{2}$.
$V_{2} = 2.176 \; L \approx 2.18 \; L$.

(N/A) We know that the formula for molar mass is $M_{2} = \frac{1000 \times w_{2} \times K_{f}}{\Delta T_{f} \times w_{1}}$.
For $AB_{2}$: $M_{AB_{2}} = \frac{1000 \times 1 \times 5.1}{2.3 \times 20} = 110.87 \ g \ mol^{-1}$.
For $AB_{4}$: $M_{AB_{4}} = \frac{1000 \times 1 \times 5.1}{1.3 \times 20} = 196.15 \ g \ mol^{-1}$.
Let the atomic masses of $A$ and $B$ be $x$ and $y$ respectively.
Then,$x + 2y = 110.87$ $(i)$ and $x + 4y = 196.15$ $(ii)$.
Subtracting $(i)$ from $(ii)$,we get $2y = 85.28$,so $y = 42.64 \ u$.
Substituting $y$ in $(i)$,$x + 2(42.64) = 110.87$,so $x = 25.59 \ u$.
The atomic masses of $A$ and $B$ are $25.59 \ u$ and $42.64 \ u$ respectively.

(0.6) Molar mass of benzene $(C_6H_6) = 78 \, g \, mol^{-1}$.
Molar mass of toluene $(C_6H_5CH_3) = 92 \, g \, mol^{-1}$.
Moles of benzene $(n_b) = \frac{80}{78} = 1.026 \, mol$.
Moles of toluene $(n_t) = \frac{100}{92} = 1.087 \, mol$.
Mole fraction of benzene $(x_b) = \frac{1.026}{1.026 + 1.087} = 0.486$.
Mole fraction of toluene $(x_t) = 1 - 0.486 = 0.514$.
Partial pressure of benzene $(p_b) = x_b \times p_b^o = 0.486 \times 50.71 = 24.645 \, mm \, Hg$.
Partial pressure of toluene $(p_t) = x_t \times p_t^o = 0.514 \times 32.06 = 16.479 \, mm \, Hg$.
Mole fraction of benzene in vapour phase $(y_b) = \frac{p_b}{p_b + p_t} = \frac{24.645}{24.645 + 16.479} = \frac{24.645}{41.124} = 0.599 \approx 0.6$.

(N/A) Ideal solutions: $n-hexane$ and $n-heptane$,or $bromoethane$ and $chloroethane$.
Azeotropes:
$1$. Minimum boiling azeotrope: $Ethanol-Water$ mixture ($95.4\%$ $ethanol$ by volume).
$2$. Maximum boiling azeotrope: $Nitric \ acid-Water$ ($68\%$ $HNO_3$ by mass).

(N/A) $1$. Boiling point: The boiling point of a liquid is the temperature at which its vapor pressure becomes equal to the atmospheric pressure $(1.013 \ bar)$.
$2$. Freezing point: The freezing point of a substance is the temperature at which the vapor pressure of the substance in its liquid phase is equal to the vapor pressure of the substance in its solid phase.

(B) Osmotic pressure $\pi = i \times C \times RT$
For $NaCl$,$i = 2$. Given $\pi_{NaCl} = 0.10 \ atm = 2 \times C_{NaCl} \times RT$. Thus,$C_{NaCl} \times RT = 0.05 \ atm$.
For glucose,$i = 1$. Given $\pi_{glucose} = 0.20 \ atm = 1 \times C_{glucose} \times RT$. Thus,$C_{glucose} \times RT = 0.20 \ atm$.
Number of moles of $NaCl$ in $1 \ L$ is $n_{NaCl} = C_{NaCl} \times 1 = \frac{0.05}{RT}$.
Number of moles of glucose in $2 \ L$ is $n_{glucose} = C_{glucose} \times 2 = \frac{0.20 \times 2}{RT} = \frac{0.40}{RT}$.
Total volume $V_{total} = 1 \ L + 2 \ L = 3 \ L$.
Total osmotic pressure $\pi_{total} = \frac{(n_{NaCl} \times i_{NaCl} + n_{glucose} \times i_{glucose}) \times RT}{V_{total}}$.
$\pi_{total} = \frac{(\frac{0.05}{RT} \times 2 + \frac{0.40}{RT} \times 1) \times RT}{3} = \frac{0.10 + 0.40}{3} = \frac{0.50}{3} \ atm$.
$\pi_{total} = 0.1666... \ atm = 166.6... \times 10^{-3} \ atm$.
Rounding to the nearest integer,$x = 167$.

(B) The dissociation of $K_{4}[Fe(CN)_{6}]$ is given by: $K_{4}[Fe(CN)_{6}] \rightleftharpoons 4K^{+} + [Fe(CN)_{6}]^{4-}$.
Initial concentration: $1 \, m$,$0$,$0$.
Final concentration: $(1 - 0.4) \, m$,$4 \times 0.4 \, m$,$0.4 \, m$.
Total effective molality $= 0.6 + 1.6 + 0.4 = 2.6 \, m$.
Since the boiling point is the same,the molality of the non-electrolytic solution must also be $2.6 \, m$.
For $18.1 \%$ by weight solution,$18.1 \, g$ of solute $A$ is present in $81.9 \, g$ of water.
Using the formula for molality: $m = \frac{w_A \times 1000}{M_A \times w_{solvent}}$.
$2.6 = \frac{18.1 \times 1000}{M_A \times 81.9}$.
$M_A = \frac{18100}{2.6 \times 81.9} \approx 85 \, u$.

(B) For acetone as solvent:
$\Delta T_{b} = i \times K_{b} \times m$
$0.17 = 1 \times 1.7 \times \frac{1.22 / M_{w}}{100 / 1000} \dots (1)$
For benzene as solvent (dimerization,$i = 0.5$):
$\Delta T_{b} = 0.5 \times 2.6 \times \frac{1.22 / M_{w}}{100 / 1000} \dots (2)$
Dividing $(2)$ by $(1)$:
$\frac{\Delta T_{b}}{0.17} = \frac{0.5 \times 2.6}{1.7} = \frac{1.3}{1.7}$
$\Delta T_{b} = \frac{1.3 \times 0.17}{1.7} = 0.13 \, ^{\circ}C$
Given $\Delta T_{b} = x \times 10^{-2} \, ^{\circ}C$,we have $0.13 = x \times 10^{-2}$,so $x = 13$.

(D) For glucose,the van't Hoff factor $i = 1$.
The elevation in boiling point is given by $\Delta T_{b} = K_{b} \cdot m_{1} = 4 \, K$,where $m_{1} = 1.5 \, m$.
So,$K_{b} \cdot 1.5 = 4 \implies K_{b} = \frac{4}{1.5}$.
The depression in freezing point is given by $\Delta T_{f} = K_{f} \cdot m_{2} = 4 \, K$,where $m_{2} = 4.5 \, m$.
So,$K_{f} \cdot 4.5 = 4 \implies K_{f} = \frac{4}{4.5}$.
The ratio $\frac{K_{b}}{K_{f}} = \frac{4 / 1.5}{4 / 4.5} = \frac{4.5}{1.5} = 3$.

(B) For elevation in boiling point: $\Delta T_{b} = K_{b} \times m_{1} = 3 \ K$ where $m_{1} = 1 \ m$. So,$K_{b} = 3 \ K \ kg \ mol^{-1}$.
For depression in freezing point: $\Delta T_{f} = K_{f} \times m_{2} = 6 \ K$ where $m_{2} = 2 \ m$. So,$K_{f} = 6 / 2 = 3 \ K \ kg \ mol^{-1}$.
The ratio $K_{b} / K_{f} = 3 / 3 = 1 / 1$.
Comparing this with $1 : X$,we get $X = 1$.

(C) The correct option is $(C)$.
When a concentrated solution of $CuSO_4$ is mixed with a dilute solution of $CuSO_4$,the process is essentially a dilution process.
During dilution,the randomness of the solute particles increases as they spread out in a larger volume,which leads to an increase in entropy $(\Delta S > 0)$.
Since the mixing of two solutions of the same solute at the same temperature involves negligible heat change (ideal mixing approximation),the enthalpy change $(\Delta H)$ is approximately zero.

(A) Given,specific heat of substance $= 0.86 \,J \,g^{-1} \,K^{-1}$.
$1 \,molal$ solution means $1 \,mole$ of solute in $1000 \,g$ of solvent (water).
Mass of solute $= 1 \,mole \times 58 \,g \,mol^{-1} = 58 \,g$.
Total mass of solution $= 1000 \,g + 58 \,g = 1058 \,g$.
In $1058 \,g$ solution,mass of solute $= 58 \,g$ and mass of water $= 1000 \,g$.
For $10 \,g$ of solution:
Mass of solute $= (58 / 1058) \times 10 \approx 0.548 \,g$.
Mass of water $= (1000 / 1058) \times 10 \approx 9.452 \,g$.
Energy required $q = (m_{solute} \times c_{solute} \times \Delta T) + (m_{water} \times c_{water} \times \Delta T)$.
$q = (0.548 \times 0.86 \times 10) + (9.452 \times 4.2 \times 10)$.
$q = 4.7128 + 396.984 = 401.6968 \,J \approx 401.7 \,J$.

(A) The vapour pressure $(V.P.)$ of a solution is always lower than that of the pure solvent due to the presence of non-volatile solute particles.
At the freezing point,the solid phase of the solvent is in equilibrium with the liquid phase of the solution. Only the solvent molecules solidify,while the solute remains in the liquid phase.
Therefore,statements $A$ and $D$ are correct.

(A) The osmotic pressure is given by $\pi = iCRT$,where $i$ is the van't Hoff factor,$C$ is the molarity,$R$ is the gas constant,and $T$ is the temperature.
For solutions at the same temperature,$\pi \propto iC$.
$A.$ $iC$ for $C_2H_5OH = 1 \times 0.500 = 0.500$; $iC$ for $KBr = 2 \times 0.25 = 0.500$. (Equal)
$B.$ $iC$ for $K_4[Fe(CN)_6] = 5 \times 0.100 = 0.500$; $iC$ for $FeSO_4(NH_4)_2SO_4 = 3 \times 0.100 = 0.300$. (Not equal)
$C.$ $iC$ for $K_4[Fe(CN)_6] = 5 \times 0.05 = 0.250$; $iC$ for $NaCl = 2 \times 0.25 = 0.500$. (Not equal)
$D.$ $iC$ for $NaCl = 2 \times 0.15 = 0.300$; $iC$ for $BaCl_2 = 3 \times 0.1 = 0.300$. (Equal)
$E.$ $iC$ for $KCl \cdot MgCl_2 \cdot 6H_2O = 3 \times 0.02 = 0.060$; $iC$ for $KCl = 2 \times 0.05 = 0.100$. (Not equal)
Only pairs $A$ and $D$ have the same osmotic pressure. Thus,the number of such pairs is $2$.

(A) Let $a$ moles of $Pb(NO_3)_2$ be added.
$Pb(NO_3)_2 \rightarrow Pb^{2+} + 2NO_3^-$
Initial moles: $a \quad 0 \quad 0$
Final moles: $0 \quad a \quad 2a$
Total particles = $3a$.
$\Delta T_b = i \times K_b \times m = 3 \times 0.5 \times a = 0.15 \Rightarrow a = 0.1 \, mol$.
Now,$Pb^{2+} + 2Cl^- \rightarrow PbCl_2(s)$.
Initial moles: $Pb^{2+} = 0.1, Cl^- = 0.2, NO_3^- = 0.2$.
Let $x$ moles of $PbCl_2$ precipitate.
Remaining moles: $Pb^{2+} = (0.1-x), Cl^- = (0.2-2x), NO_3^- = 0.2$.
Total moles of solute particles = $(0.1-x) + (0.2-2x) + 0.2 = 0.5-3x$.
$\Delta T_f = K_f \times m = 1.8 \times (0.5-3x) = 0.8$.
$0.9 - 5.4x = 0.8$ $\Rightarrow 5.4x = 0.1$ $\Rightarrow x = 0.1/5.4 = 1/54$.
$[Pb^{2+}] = 0.1 - 1/54 = 4.4/54 \approx 0.0815$.
$[Cl^-] = 0.2 - 2/54 = 8.8/54 \approx 0.163$.
$K_{sp} = [Pb^{2+}][Cl^-]^2 = (4.4/54) \times (8.8/54)^2 \approx 13 \times 10^{-6}$.

(A) van't Hoff factor,$i = \frac{\text{Normal molar mass}}{\text{Abnormal molar mass}}$,which corresponds to $III$.
$(B)$ $k_{f}$ is the Cryoscopic constant (molal depression constant),which corresponds to $I$.
$(C)$ Solutions with the same osmotic pressure are known as Isotonic solutions,which corresponds to $II$.
$(D)$ Azeotropes are binary mixtures having the same composition in liquid and vapour phases,which corresponds to $IV$.
Therefore,the correct matching is $A-III, B-I, C-II, D-IV$.

(D) Incorrect. $\Delta T_b = i \times K_b \times m$. For $NaCl$,$i=2$,and for urea,$i=1$. Thus,elevation is not the same.
$(B)$ Correct. Azeotropic mixtures boil at a constant temperature and have the same composition in both liquid and vapour phases.
$(C)$ Incorrect. Osmosis always takes place from a hypotonic solution (lower concentration) to a hypertonic solution (higher concentration) through a semi-permeable membrane.
$(D)$ Correct. Molarity $M = \frac{\% \times 10 \times d}{M_{solute}} = \frac{32 \times 10 \times 1.26}{98} \approx 4.11 \ M$,which is approximately $4.09 \ M$.
$(E)$ Incorrect. When $KI$ is added to $AgNO_3$ (excess $AgNO_3$),$AgI$ adsorbs $Ag^+$ ions to form a positively charged sol $(AgI/Ag^+)$.
Therefore,only statements $(B)$ and $(D)$ are correct.

(A) The reaction is: $KCl + AgNO_3 \rightarrow AgCl + KNO_3$.
At the equivalence point,the millimoles of $KCl$ equal the millimoles of $AgNO_3$:
$n(KCl) = 20 \ mL \times 1 \ M = 20 \ mmol = 0.02 \ mol$.
Mass of the solution $= 25 \ mL \times 1 \ g \ mL^{-1} = 25 \ g$.
Mass of $KCl$ solute $= 0.02 \ mol \times 74.5 \ g \ mol^{-1} = 1.49 \ g$.
Mass of solvent $= 25 \ g - 1.49 \ g = 23.51 \ g = 0.02351 \ kg$.
Molality $(m) = \frac{0.02 \ mol}{0.02351 \ kg} \approx 0.8507 \ mol \ kg^{-1}$.
Since $KCl$ undergoes $100 \%$ ionization,the van't Hoff factor $i = 2$.
Depression in freezing point $\Delta T_f = i \times K_f \times m = 2 \times 2.0 \times 0.8507 = 3.4028 \ K$.
The nearest integer is $3$.

(B) $1$. Calculate the molar mass of acetic acid $(CH_3COOH)$: $12 + 3(1) + 12 + 16 + 16 + 1 = 60 \ g \ mol^{-1}$.
$2$. Calculate the van't Hoff factor $(i)$ for $20\%$ dissociation $(\alpha = 0.2)$: $i = 1 + \alpha(n-1) = 1 + 0.2(2-1) = 1.2$.
$3$. Calculate the molality $(m)$ of the solution: $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{5/60}{0.5} = \frac{1}{6} \times 2 = 0.333 \ mol \ kg^{-1}$.
$4$. Calculate the depression in freezing point $(\Delta T_f)$: $\Delta T_f = i \times K_f \times m = 1.2 \times 1.86 \times \frac{5 \times 1000}{60 \times 500} = 1.2 \times 1.86 \times 0.1666 = 0.372 \ { }^{\circ}C$.
$5$. Express in the required format: $0.372 \ { }^{\circ}C = 372 \times 10^{-3} \ { }^{\circ}C$.

(B) The depression in freezing point is given by $\Delta T_f = i \cdot K_f \cdot m$.
$1.$ For $180 \ g$ of acetic acid ($CH_3COOH$,molar mass $= 60 \ g/mol$) in benzene,$m = 3 \ mol/kg$ (assuming $1 \ kg$ solvent). Acetic acid dimerizes in benzene,so $i \approx 0.5$. $\Delta T_f = 0.5 \times 5.12 \times 3 = 7.68 \ K$.
$2.$ For $180 \ g$ of benzoic acid ($C_6H_5COOH$,molar mass $= 122 \ g/mol$) in benzene,$m = 180/122 \approx 1.47 \ mol/kg$. $i \approx 0.5$. $\Delta T_f = 0.5 \times 5.12 \times 1.47 \approx 3.76 \ K$.
$3.$ For $180 \ g$ of acetic acid in water,$m = 3 \ mol/kg$. Acetic acid is a weak electrolyte,$i \approx 1 + \alpha \approx 1$. $\Delta T_f = 1 \times 1.86 \times 3 = 5.58 \ K$.
$4.$ For $180 \ g$ of glucose ($C_6H_{12}O_6$,molar mass $= 180 \ g/mol$) in water,$m = 1 \ mol/kg$. $i = 1$. $\Delta T_f = 1 \times 1.86 \times 1 = 1.86 \ K$.
Comparing all,acetic acid in benzene shows the highest $\Delta T_f$.

(D) Given: $\Delta T_b = 2 \ K$,$K_b = 0.52 \ K \ kg \ mol^{-1}$,$W_{\text{solvent}} = 100 \ g$,$M_{\text{solvent}} = 18 \ g \ mol^{-1}$,$P^{\circ} = 760 \ mm \ Hg$.
Step $1$: Calculate molality $(m)$ using $\Delta T_b = K_b \times m$.
$2 = 0.52 \times m \implies m = \frac{2}{0.52} \approx 3.846 \ mol \ kg^{-1}$.
Step $2$: Use Raoult's law for dilute solutions: $\frac{P^{\circ} - P_s}{P^{\circ}} = \frac{n_{\text{solute}}}{n_{\text{solvent}}}$.
Since $m = \frac{n_{\text{solute}}}{W_{\text{solvent}}(kg)}$,we have $n_{\text{solute}} = m \times \frac{W_{\text{solvent}}}{1000}$.
$n_{\text{solvent}} = \frac{100}{18} = 5.556 \ mol$.
Step $3$: Calculate relative lowering of vapour pressure.
$\frac{\Delta P}{P^{\circ}} = \frac{m \times W_{\text{solvent}}}{1000} \times \frac{1}{n_{\text{solvent}}} = \frac{m \times M_{\text{solvent}}}{1000}$.
$\Delta P = 760 \times \frac{3.846 \times 18}{1000} = 760 \times 0.069228 = 52.613 \ mm \ Hg$.
Step $4$: Calculate vapour pressure of solution $(P_s)$.
$P_s = P^{\circ} - \Delta P = 760 - 52.613 = 707.387 \ mm \ Hg$.
Rounding to the nearest integer,we get $707 \ mm \ Hg$.