Mix Examples of Solutions Questions in English

(B) Beckmann thermometer is a sensitive thermometer used to measure small changes in temperature,typically in the range of $5 \ ^\circ C$ to $6 \ ^\circ C$. It is commonly used in experiments like freezing point depression or boiling point elevation.

(C) Sugar $(C_{12}H_{22}O_{11})$ is a non-electrolyte and does not dissociate into ions in water.
When sugar dissolves in water,the sugar molecules disperse throughout the water,occupying the intermolecular spaces between the water molecules.
Therefore,the solution contains free sugar molecules and free water molecules.

(D) Given mole ratio of pentane $(n_p)$ to hexane $(n_h)$ is $1:4$.
Mole fraction of pentane $(x_p)$ = $\frac{1}{1+4} = 0.2$.
Mole fraction of hexane $(x_h)$ = $\frac{4}{1+4} = 0.8$.
Total vapour pressure $(P_T)$ = $P_p^0 x_p + P_h^0 x_h$.
$P_T = (440 \times 0.2) + (120 \times 0.8) = 88 + 96 = 184 \ mm \ Hg$.
According to Dalton's law,the mole fraction of pentane in the vapour phase $(y_p)$ is given by $y_p = \frac{P_p^0 x_p}{P_T}$.
$y_p = \frac{88}{184} \approx 0.478$.

(D) Colligative properties depend only on the number of solute particles in a solution,not on their chemical nature.
$1$. Addition of non-volatile solutes to a solvent leads to a decrease in vapour pressure and an increase in boiling point (elevation of boiling point).
$2$. In option $C$,the addition of non-volatile naphthalene to benzene decreases the vapour pressure of benzene,which is a colligative property.
$3$. However,the second part of option $C$ (addition of toluene to benzene) involves two volatile liquids,which does not strictly follow the standard definition of colligative properties as they are typically defined for non-volatile solutes.
$4$. Since options $A$ and $B$ involve volatile solutes (ethanol and nitric acid),they do not represent standard colligative effects.
$5$. Therefore,none of the provided statements correctly describe a pure colligative effect based on standard definitions.

(D) The osmotic pressure $(\pi)$ of a solution is given by the van't Hoff equation: $\pi = CRT$,where $C$ is the molar concentration,$R$ is the gas constant,and $T$ is the absolute temperature.
Thus,$\pi$ is directly proportional to the molecular concentration $(C)$ and the absolute temperature $(T)$.
Additionally,osmotic pressure is related to the relative lowering of vapour pressure by the expression: $\pi = \left( \frac{P^0 - P_s}{P^0} \right) \times \frac{dRT}{M}$,where $d$ is the density of the solution and $M$ is the molar mass of the solute.
Therefore,all the given statements are correct.

(D) Osmotic pressure $(\pi)$ is given by the formula $\pi = CRT$,where $C$ is the molar concentration,$R$ is the gas constant,and $T$ is the temperature. For the same temperature,solutions with the same molarity $(C)$ will have the same osmotic pressure.
$1$. For $0.1 \, M$ glucose: $C = 0.1 \, M$.
$2$. For $0.6 \, g$ urea (molar mass $= 60 \, g/mol$) in $100 \, mL$ solution: $C = \frac{0.6/60}{0.1} = \frac{0.01}{0.1} = 0.1 \, M$.
$3$. For $1.0 \, g$ of solute $X$ (molar mass $= 200 \, g/mol$) in $50 \, mL$ solution: $C = \frac{1.0/200}{0.05} = \frac{0.005}{0.05} = 0.1 \, M$.
Since all solutions have a molarity of $0.1 \, M$,they all produce the same osmotic pressure.

(D) The correct answer is $(d)$.
Freezing point depression $(\Delta T_f)$ is a colligative property given by $\Delta T_f = K_f \times m$,where $K_f$ is the cryoscopic constant of the solvent and $m$ is the molality.
Since $K_f$ is a characteristic property of the solvent,it varies from one solvent to another.
Therefore,even if the molality $(m)$ is the same,the freezing point depression will differ if the solvents are different.

(B) Given that the boiling point of the aqueous solution is $100.52 \, ^\circ C$,the elevation in boiling point is $\Delta T_b = 100.52 \, ^\circ C - 100 \, ^\circ C = 0.52 \, ^\circ C$.
For water,the molal elevation constant $K_b = 0.52 \, K \cdot kg \cdot mol^{-1}$.
Using the formula $\Delta T_b = K_b \times m$,we get $0.52 = 0.52 \times m$,which implies the molality $m = 1 \, mol \cdot kg^{-1}$.
The depression in freezing point is given by $\Delta T_f = K_f \times m$.
For water,the molal depression constant $K_f = 1.86 \, K \cdot kg \cdot mol^{-1}$.
Thus,$\Delta T_f = 1.86 \times 1 = 1.86 \, ^\circ C$.
The freezing point of the solution $T_f = 0 \, ^\circ C - \Delta T_f = 0 \, ^\circ C - 1.86 \, ^\circ C = - 1.86 \, ^\circ C$.

(B) The depression in freezing point is given by $\Delta T_f = T_f^{\circ} - T_f = 0 - (-0.186) = 0.186 \ ^{\circ}C$.
Using the formula $\Delta T_f = K_f \times m$,we have $0.186 = 1.86 \times m$.
Therefore,the molality $m = \frac{0.186}{1.86} = 0.1 \ mol \ kg^{-1}$.
Now,calculate the elevation in boiling point using $\Delta T_b = K_b \times m$.
$\Delta T_b = 0.521 \times 0.1 = 0.0521 \ K$.

(D) The freezing point of a solution is the temperature at which the vapour pressure of the solution becomes equal to the vapour pressure of the pure solid solvent.
At the freezing point,only the solvent molecules solidify,while the solute molecules remain in the liquid phase.
Furthermore,the addition of a non-volatile solute lowers the vapour pressure of the solvent,which leads to a depression in the freezing point.
Therefore,both statements $A$ and $B$ are correct.

(A, B, C, D) The depression in freezing point is a colligative property,which depends on the number of solute particles in the solution. The formula is $\Delta T_f = i \times K_f \times m$. Since all the given solutions have the same molality $(0.1 \ m)$ and are non-electrolytes,they all have the same van't Hoff factor $(i = 1)$. Therefore,they all exhibit the same depression in freezing point and consequently have the same freezing point. However,if the question implies a comparison where one might be different,in this specific set,all options are non-electrolytes with equal concentration,meaning they are identical in their effect on freezing point.

(B) When ethanol $(C_2H_5OH)$ dissolves in water $(H_2O)$,the process is exothermic,meaning there is an emission of heat.
Additionally,due to the formation of strong hydrogen bonds between ethanol and water molecules,the total volume of the solution is less than the sum of the individual volumes of ethanol and water,resulting in a contraction in volume.
Therefore,the correct option is $(B)$.

(A) At atmospheric pressure,the boiling point of acetone is $56^{\circ}C$.
Acetone is a polar molecule,which leads to dipole-dipole interactions between its molecules in the liquid phase.
These interactions are weaker than the hydrogen bonds found in water,resulting in a lower boiling point for acetone compared to water.

(A) Urea $(NH_2CONH_2)$ is a non-electrolyte and does not dissociate into ions in water. Therefore,its aqueous solution is neutral in nature.

(D) $1$. Raoult's law states that the partial vapor pressure of each component in a solution is directly proportional to its mole fraction in the solution. This is correct.
$2$. The osmotic pressure formula is $\pi = MRT$,where $M$ is the molarity. This is correct.
$3$. Osmotic pressure is a colligative property depending on the van't Hoff factor $(i)$. For $0.01 \ M$ solutions: $BaCl_2$ $(i=3)$,$KCl$ $(i=2)$,$CH_3COOH$ $(i \approx 1)$,and sucrose $(i=1)$. The order $BaCl_2 > KCl > CH_3COOH > \text{sucrose}$ is correct.
$4$. Depression in freezing point $(\Delta T_f = K_f \cdot m)$ depends on the cryoscopic constant $(K_f)$ of the solvent. Since $K_f$ is a characteristic property of the solvent,different solvents will have different $K_f$ values,leading to different depressions in freezing point even if the molality $(m)$ is the same. Thus,statement $D$ is incorrect.

(A) $1$. Calculate the molality $(m)$ of the solution using the elevation in boiling point formula: $\Delta T_b = K_b \times m$.
$2$. Given $\Delta T_b = 100.18 - 100 = 0.18 \ K$ and $K_b = 0.512 \ K \ kg \ mol^{-1}$.
$3$. $m = \frac{\Delta T_b}{K_b} = \frac{0.18}{0.512} \approx 0.3516 \ mol \ kg^{-1}$.
$4$. Calculate the depression in freezing point: $\Delta T_f = K_f \times m$.
$5$. $\Delta T_f = 1.86 \times 0.3516 \approx 0.654 \ K$.
$6$. The freezing point of the solution is $T_f = T_f^\circ - \Delta T_f = 273.15 \ K - 0.654 \ K = 272.496 \ K \approx 272.50 \ K$.
$7$. Based on the calculation,the closest option is $272.73 \ K$ (noting potential rounding differences in standard problems).

(B) The depression in freezing point is $\Delta T_f = 0 - (-0.186) = 0.186 \ ^oC$.
We know that $\Delta T_f = K_f \times m$ and $\Delta T_b = K_b \times m$.
Therefore,$\frac{\Delta T_f}{\Delta T_b} = \frac{K_f}{K_b}$.
Substituting the values: $\frac{0.186}{\Delta T_b} = \frac{1.86}{0.512}$.
Solving for $\Delta T_b$: $\Delta T_b = \frac{0.186 \times 0.512}{1.86} = 0.0512 \ ^oC$.
The boiling point of the solution is $(T_b)_{solution} = (T_b)_{solvent} + \Delta T_b = 100 + 0.0512 = 100.0512 \ ^oC$.

(D) The sum of mole fractions of solute and solvent is always $1$,i.e.,$\chi_{\text{solute}} + \chi_{\text{solvent}} = 1$.
If the mole fraction of the solvent $(\chi_{\text{solvent}})$ decreases,the mole fraction of the solute $(\chi_{\text{solute}})$ must increase.
An increase in the concentration of the solute leads to an elevation in the boiling point $(\Delta T_b = K_b \times m)$ and an increase in osmotic pressure $(\pi = CRT)$.
Therefore,all the given statements are correct.

(A) The elevation in boiling point is given by $\Delta T_b = \frac{1000 \times K_b \times w_2}{M_2 \times w_1}$.
Given $\Delta T_b = 2 \ K$,$K_b = 0.76 \ K \ kg \ mol^{-1}$,$w_2 = 2.5 \ g$,$w_1 = 100 \ g$.
$2 = \frac{1000 \times 0.76 \times 2.5}{M_2 \times 100}$ $\Rightarrow M_2 = \frac{1000 \times 0.76 \times 2.5}{200} = 9.5 \ g \ mol^{-1}$.
Using Raoult's law for relative lowering of vapor pressure: $\frac{P^0 - P_s}{P^0} = \frac{n_2}{n_1} = \frac{w_2 / M_2}{w_1 / M_1}$.
Here $P^0 = 760 \ mm \ Hg$,$M_1 = 18 \ g \ mol^{-1}$.
$\frac{760 - P_s}{760} = \frac{2.5 / 9.5}{100 / 18} = \frac{2.5 \times 18}{9.5 \times 100} = \frac{45}{950} \approx 0.04737$.
$760 - P_s = 760 \times 0.04737 = 36$.
$P_s = 760 - 36 = 724 \ mm \ Hg$.

(D) When the mixture boils at $80^o C$ and $1 \ atm$ pressure,the total vapor pressure $P_s$ is equal to the external pressure,i.e.,$P_s = 760 \ mm \ Hg$.
According to Raoult's Law: $P_s = P_A^0 X_A + P_B^0 X_B$.
Since $X_A + X_B = 1$,we have $X_B = 1 - X_A$.
Substituting the values: $760 = 520 X_A + 1000(1 - X_A)$.
$760 = 520 X_A + 1000 - 1000 X_A$.
$480 X_A = 240$.
$X_A = 0.50$.
Therefore,the mole percentage of $A$ is $0.50 \times 100 = 50 \%$.

(B) Statement $1$ describes an endothermic process where the dissolution of a solute (like ammonium nitrate) in water absorbs heat from the surroundings,lowering the temperature. This is a common application of cooling packs.
Statement $2$ describes the colligative property known as the depression of freezing point,which states that the freezing point of a solvent decreases when a non-volatile solute is added.
While both statements are scientifically correct,the cooling effect in a cold pack is primarily due to the endothermic nature of the dissolution process (enthalpy of solution),not directly due to the depression of the freezing point of the solvent. Therefore,Statement $2$ is not the explanation for Statement $1$.

(A) The depression in freezing point is given by $\Delta T_f = T_f^o - T_f$. Assuming the solvent is water,$T_f^o = 273.15 \ K$ (or $0^\circ C$).
Given the freezing point of the solution is $1.86 \ K$ (assuming the decrease is $1.86 \ K$ relative to the solvent),$\Delta T_f = 1.86 \ K$.
Using the formula $\Delta T_f = K_f \times m$,we have $1.86 = 1.86 \times m$,which gives molality $m = 1 \ m$.
Now,the elevation in boiling point is $\Delta T_b = K_b \times m$.
Substituting the values,$\Delta T_b = 0.512 \times 1 = 0.512 \ K$.

(D) Urea is a non-electrolyte,so its van't Hoff factor $(i)$ is $1$.
$Al_2(SO_4)_3$ is an electrolyte that dissociates as $Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-}$,so its van't Hoff factor $(i)$ is $5$.
Colligative properties like elevation in boiling point $(\Delta T_b = i K_b m)$ and depression in freezing point $(\Delta T_f = i K_f m)$ are directly proportional to $i$.
Since $i_{Al_2(SO_4)_3} > i_{urea}$,$Al_2(SO_4)_3$ will have a higher elevation in boiling point and a higher depression in freezing point.
Lower depression in freezing point means a higher freezing point for urea compared to $Al_2(SO_4)_3$.
Lower solute particles mean higher vapor pressure for urea compared to $Al_2(SO_4)_3$.
Thus,statements $a, b, c,$ and $d$ are correct.

(A) The formula for depression in freezing point is $\Delta T_f = i \times K_f \times m$.
For $NaCl$,$i = 2$ (since $NaCl \rightarrow Na^+ + Cl^-$).
Given $\Delta T_{f(NaCl)} = 0.37^o C$ and $m = 0.01 \, m$.
$0.37 = 2 \times K_f \times 0.01 \implies K_f = \frac{0.37}{0.02} = 18.5$.
For urea,$i = 1$ (non-electrolyte) and $m = 0.02 \, m$.
$\Delta T_{f(urea)} = 1 \times K_f \times 0.02 = 1 \times 18.5 \times 0.02 = 0.37^o C$.

(A) Proof spirit is defined as a mixture containing $57.1 \%$ of ethyl alcohol by volume and $42.9 \%$ of water by volume.
Therefore,the correct option is $A$.

(C) Given: $W_{B} = 6.5\, g$,$W_{A} = 100\, g$,$p_{s} = 732\, mm$,$K_{b} = 0.52$,$T_{b}^{o} = 100\, ^oC$,$p^{o} = 760\, mm$.
Using Raoult's law for relative lowering of vapour pressure: $\frac{p^{o} - p_{s}}{p^{o}} = \frac{n_{2}}{n_{1}}$.
$\frac{760 - 732}{760} = \frac{n_{2}}{100 / 18}$.
$n_{2} = \frac{28 \times 100}{760 \times 18} \approx 0.2046\, mol$.
Now,calculate elevation in boiling point: $\Delta T_{b} = K_{b} \times m = K_{b} \times \frac{n_{2} \times 1000}{W_{A} (g)}$.
$\Delta T_{b} = \frac{0.52 \times 0.2046 \times 1000}{100} = 1.06\, ^oC$.
Boiling point of solution $T_{b} = T_{b}^{o} + \Delta T_{b} = 100 + 1.06 = 101.06\, ^oC$.

(B) The total vapour pressure of a binary solution is given by Raoult's Law: $P_s = P_A^0 X_A + P_B^0 X_B$.
Since $X_A + X_B = 1$,we have $X_B = 1 - X_A$.
Substituting this into the equation: $P_s = P_A^0 X_A + P_B^0 (1 - X_A) = (P_A^0 - P_B^0) X_A + P_B^0$.
Comparing this with the given equation $P_s = 120 X_A + 138$:
$1$. When $X_A = 0$,$P_s = P_B^0 = 138 \, \text{torr}$.
$2$. When $X_A = 1$,$P_s = P_A^0 = 120(1) + 138 = 258 \, \text{torr}$.
Thus,$P_B^0 = 138 \, \text{torr}$ and $P_A^0 = 258 \, \text{torr}$.

(B) The elevation in boiling point is given by $\Delta T_b = i \times K_b \times m$.
For an alkali metal chloride $(MCl)$,the van't Hoff factor $i = 2$.
Given $\Delta T_b = 100.051 - 100 = 0.051 \ ^\circ C$,$K_b = 0.51 \ K \ kg \ mol^{-1}$,mass of solute $w = 1.685 \ g$,and mass of solvent $W = 200 \ g$.
$0.051 = 2 \times 0.51 \times \frac{1.685}{M} \times \frac{1000}{200}$.
$0.051 = 1.02 \times \frac{1.685}{M} \times 5$.
$0.051 = \frac{8.425}{M} \implies M = \frac{8.425}{0.051} \approx 165.2$.
Considering the given atomic masses,$CsCl$ $(133 + 35.5 = 168.5)$ is the closest match.
$CsCl$ crystallizes in a $bcc$ structure where the edge length $a$ is related to the radii by $\frac{\sqrt{3} a}{2} = r^+ + r^-$.
Given $r^+ = 1.70 \ \mathring{A}$ and $r^- = 1.80 \ \mathring{A}$,$r^+ + r^- = 3.50 \ \mathring{A}$.
$\frac{\sqrt{3} a}{2} = 3.5 \implies a = \frac{7}{\sqrt{3}} \ \mathring{A}$.

(A) $1$. The statement in option $A$ is false because in solutions showing negative deviation from Raoult's law,the volume of the solution is less than the sum of the volumes of the pure components $(\Delta V_{mix} < 0)$.
$2$. For an ideal solution,$\Delta G_{mix} = \Delta H_{mix} - T\Delta S_{mix}$. Since $\Delta H_{mix} = 0$ and $\Delta S_{mix} > 0$,$\Delta G_{mix}$ is always negative.
$3$. Ideal solutions follow Raoult's law at all concentrations and temperatures,thus they cannot form azeotropes (constant boiling mixtures).
$4$. Perfectly ideal solutions are rare; most real solutions show deviations,making ideality an exception.

(B) The dissociation of $PbCl_2$ is given by: $PbCl_{2(s)} \leftrightarrow Pb^{2+}_{(aq)} + 2Cl^-_{(aq)}$
The solubility product expression is: $K_{sp} = [Pb^{2+}][Cl^-]^2 = (s)(2s)^2 = 4s^3$
Given $K_{sp} = 4 \times 10^{-6}$,we have $4s^3 = 4 \times 10^{-6}$,which gives $s = 10^{-2} \ mol \ L^{-1}$.
For $PbCl_2$,the Van't Hoff factor $i = 3$ (one $Pb^{2+}$ and two $Cl^-$ ions).
The depression in freezing point is calculated as: $\Delta T_f = i \times K_f \times m$.
Assuming $m \approx s = 10^{-2} \ mol \ kg^{-1}$,we get $\Delta T_f = 3 \times 2 \times 10^{-2} = 0.06 \ K$.
Since the freezing point of pure water is $0 \ ^oC$,the freezing point of the solution is $0 - 0.06 = -0.06 \ ^oC$.

(A) Consider $100 \, g$ of the solution. The mass of $HA = 6 \, g$ and the mass of urea = $4 \, g$.
The mass of water = $100 - (6 + 4) = 90 \, g$.
For $HA$,$n_1 = \frac{6}{60} = 0.1 \, mol$. For urea,$n_2 = \frac{4}{60} = 0.0667 \, mol$.
Using the formula $\Delta T_f = K_f \times \text{molality} \times i_{total}$,where $i_{total} = \frac{i_1 n_1 + i_2 n_2}{n_1 + n_2}$.
$\Delta T_f = K_f \times \frac{(i_1 n_1 + i_2 n_2) \times 1000}{W_{solvent(g)}}$.
$\frac{13.6}{3} = 1.8 \times \frac{((1+\alpha) \times 0.1 + 1 \times 0.0667) \times 1000}{90}$.
$\frac{13.6}{3} = 20 \times (0.1 + 0.1\alpha + 0.0667) = 20 \times (0.1667 + 0.1\alpha)$.
$4.533 = 3.334 + 2\alpha$ $\Rightarrow 2\alpha = 1.199$ $\Rightarrow \alpha \approx 0.6$.
$K_a = \frac{C\alpha^2}{1-\alpha}$. Here $C = \frac{0.1 \, mol}{0.09 \, L} = 1.11 \, M$.
$K_a = \frac{1.11 \times (0.6)^2}{1-0.6} = \frac{1.11 \times 0.36}{0.4} \approx 1$.

(B) Given: $\Delta T_b = 101.08 - 100 = 1.08\,^{\circ}C$ and $\Delta T_f = 0 - (-1.80) = 1.80\,^{\circ}C$.
For $AB$,at the boiling point,it is $100\%$ ionised,so the van't Hoff factor $i_b = 2$.
The formulas for elevation in boiling point and depression in freezing point are $\Delta T_b = i_b K_b m$ and $\Delta T_f = i_f K_f m$.
Dividing the two equations: $\frac{\Delta T_b}{\Delta T_f} = \frac{i_b K_b m}{i_f K_f m} = \frac{i_b}{i_f} \times \frac{K_b}{K_f}$.
Substituting the values: $\frac{1.08}{1.80} = \frac{2}{i_f} \times 0.3$.
$0.6 = \frac{0.6}{i_f}$,which gives $i_f = 1$.
Since the van't Hoff factor $i_f = 1$,the solute $AB$ behaves as a non-electrolyte at the freezing point of the solution.

(C) Total mass of final solution $= 100 \ g + 100 \ g = 200 \ g$.
Density of final solution $= 1.25 \ g/mL$.
Volume of final solution $= \frac{\text{mass}}{\text{density}} = \frac{200 \ g}{1.25 \ g/mL} = 160 \ mL$.
Total mass of $NaOH$ in the mixture $= (20 \% \text{ of } 100 \ g) + (40 \% \text{ of } 100 \ g) = 20 \ g + 40 \ g = 60 \ g$.
$\% \ w/v = \frac{\text{mass of solute (g)}}{\text{volume of solution (mL)}} \times 100 = \frac{60 \ g}{160 \ mL} \times 100 = 37.5 \%$.
Therefore,the correct option is $C$.

(D) $1$. In an ideal solution,the intermolecular forces between solute-solute,solvent-solvent,and solute-solvent are identical. Thus,the heat of vaporisation remains similar as the same forces are overcome.
$2$. The entropy of a solution is higher than that of a pure solvent due to increased disorder. Therefore,the entropy change $(\Delta S)$ between the solution and its vapour is smaller than that between the pure solvent and its vapour.
$3$. Due to the presence of a non-volatile solute,the vapour pressure of the solution decreases,which leads to an elevation in the boiling point compared to the pure solvent.
$4$. Since all statements are scientifically accurate,the correct option is $D$.

(C) The relative lowering of vapour pressure is given by $\frac{P^0 - P_s}{P^0} = i \times X_{solute}$.
For $Ag_3PO_4$,the van't Hoff factor $i = 4$ $(Ag_3PO_4 \rightarrow 3Ag^+ + PO_4^{3-})$.
$\frac{760 - 750}{760} = 4 \times X_{solute}$.
$\frac{10}{760} = 4 \times X_{solute} \Rightarrow X_{solute} = \frac{1}{304}$.
Since $X_{solute} = \frac{n_{solute}}{n_{solute} + n_{solvent}} \approx \frac{n_{solute}}{n_{solvent}}$,we have $\frac{n_{solute}}{n_{solvent}} = \frac{1}{304}$.
For $1 \ L$ of water,$n_{solvent} = \frac{1000 \ g}{18 \ g/mol} = 55.55 \ mol$.
$n_{solute} = \frac{55.55}{304} \approx 0.1827 \ mol/L$.
$0.1827 \approx \frac{10}{54}$.

(C) Vapour pressure expected by Raoult's law is calculated as:
$P_{\text{expected}} = P_A^o X_A + P_B^o X_B = 45 \times \frac{1}{1+2} + 36 \times \frac{2}{1+2}$
$P_{\text{expected}} = 45 \times \frac{1}{3} + 36 \times \frac{2}{3} = 15 + 24 = 39 \, \text{torr}$.
Since the observed vapour pressure $(42 \, \text{torr})$ is greater than the expected vapour pressure $(39 \, \text{torr})$,the solution exhibits positive deviation from Raoult's law.
Solutions showing positive deviation form minimum boiling azeotropes at specific compositions.

(A) The elevation in boiling point is given by $\Delta T_b = K_b \times m$.
Given $\Delta T_b = 2 \ ^\circ C$,$K_b = 0.76 \ K \ kg \ mol^{-1}$,mass of solute $w_2 = 2.5 \ g$,and mass of solvent $w_1 = 100 \ g$.
$2 = 0.76 \times \frac{2.5 \times 1000}{M_2 \times 100}$.
$M_2 = \frac{0.76 \times 2.5 \times 10}{2} = 9.5 \ g \ mol^{-1}$.
Using Raoult's law for relative lowering of vapour pressure: $\frac{P^\circ - P_s}{P^\circ} = \frac{n_2}{n_1} = \frac{w_2 \times M_1}{M_2 \times w_1}$.
Here $P^\circ = 760 \ mm \ Hg$ and $M_1 = 18 \ g \ mol^{-1}$.
$\frac{760 - P_s}{760} = \frac{2.5 \times 18}{9.5 \times 100} = \frac{45}{950} \approx 0.04737$.
$760 - P_s = 760 \times 0.04737 = 36$.
$P_s = 760 - 36 = 724 \ mm \ Hg$.

(D) According to Raoult's Law,$P_{total} = P_A^0 x_A + P_B^0 x_B$.
For the first case: $n_A = 1, n_B = 1$. Mole fractions are $x_A = 0.5, x_B = 0.5$. $P_{total} = 100 \, \text{mm Hg}$.
$0.5 P_A^0 + 0.5 P_B^0 = 100 \implies P_A^0 + P_B^0 = 200 \quad (1)$.
For the second case: $n_A = 1, n_B = 1 + 2 = 3$. Total moles = $4$. Mole fractions are $x_A = 0.25, x_B = 0.75$. $P_{total} = 100 - 20 = 80 \, \text{mm Hg}$.
$0.25 P_A^0 + 0.75 P_B^0 = 80 \implies P_A^0 + 3 P_B^0 = 320 \quad (2)$.
Subtracting $(1)$ from $(2)$: $(P_A^0 + 3 P_B^0) - (P_A^0 + P_B^0) = 320 - 200 \implies 2 P_B^0 = 120 \implies P_B^0 = 60 \, \text{mm Hg}$.
Substituting $P_B^0 = 60$ into $(1)$: $P_A^0 + 60 = 200 \implies P_A^0 = 140 \, \text{mm Hg}$.
Thus,the vapour pressures are $140 \, \text{mm Hg}$ and $60 \, \text{mm Hg}$.

(D) The depression in freezing point is given by $\Delta T_f = T_f^\circ - T_f = 0 \ ^oC - (-2.55 \ ^oC) = 2.55 \ K$.
Since $\Delta T_f = K_f \times m$,the molality $m = \frac{\Delta T_f}{K_f} = \frac{2.55}{1.86} \ mol \ kg^{-1} \approx 1.371 \ mol \ kg^{-1}$.
The elevation in boiling point is given by $\Delta T_b = K_b \times m$.
$\Delta T_b = 0.52 \ K \ kg \ mol^{-1} \times 1.371 \ mol \ kg^{-1} \approx 0.713 \ K$.
The boiling point of the solution is $T_b = T_b^\circ + \Delta T_b = 100 \ ^oC + 0.713 \ ^oC = 100.713 \ ^oC \approx 100.7 \ ^oC$.

(D) $1$. Raoult's law states that the partial vapour pressure of each component in a solution is directly proportional to its mole fraction,which is a correct statement.
$2$. The osmotic pressure formula is $\pi = iCRT$,where $i$ is the van't Hoff factor,$C$ is molarity,$R$ is the gas constant,and $T$ is temperature. This is correct.
$3$. Osmotic pressure is a colligative property depending on the number of particles. For $0.02 \ M$ solutions: $SrCl_2$ $(i=3)$,$NaCl$ $(i=2)$,$HCOOH$ $(i \approx 1)$,$Sucrose$ $(i=1)$. Thus,$SrCl_2 > NaCl > HCOOH > Sucrose$ is correct.
$4$. Freezing point depression is given by $\Delta T_f = K_f \times m$. Since $K_f$ (cryoscopic constant) is a characteristic property of the solvent,different solvents have different $K_f$ values. Therefore,two solutions of the same molality in different solvents will have different freezing point depressions. This statement is false.

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

(D) Statement $D$ is false.
Freezing point depression is given by $\Delta T_f = K_f \times m$.
Here,$K_f$ is the molal depression constant (cryoscopic constant),which depends on the nature of the solvent.
Since the solvents are different,their $K_f$ values will be different,leading to different freezing point depressions even if the molality $m$ is the same.

(A) Henry's law states that $p = K_H \cdot x$,where $p$ is the partial pressure of the gas,$K_H$ is Henry's law constant,and $x$ is the mole fraction of the gas in the solution.
$1$. $K_H$ is a function of the nature of the gas. For different gases at the same temperature,$K_H$ values differ.
$2$. Henry's law is applicable to gases that do not react chemically with the solvent. $HCl_{(g)}$ is highly soluble in water and reacts to form $H_3O^+$ and $Cl^-$,thus it does not follow Henry's law.
$3$. Henry's law is valid for gases with low solubility.
$4$. $K_H$ increases with an increase in temperature,which explains why gases become less soluble at higher temperatures.
$5$. Comparing $O_2$ and $H_2$,$K_H$ for $O_2$ is actually lower than $H_2$ at the same temperature,making option $A$ the incorrect statement.

(B) The depression in freezing point is given by $\Delta T_f = i \times K_f \times m$.
Since $K_f$ and $m$ are constant,$\Delta T_f \propto i$.
Freezing point $T_f = T_f^0 - \Delta T_f = 273 - (i \times K_f \times 0.1)$.
For sucrose,$i = 1$,so $\Delta T_f = 273 - 272 = 1 \ K$. Thus,$K_f \times 0.1 = 1$.
Now calculate $i$ for each solution:
$a$. $BaCl_2 \rightarrow Ba^{2+} + 2Cl^-$,$i = 3$. $\Delta T_f = 3 \times 1 = 3 \ K$. $T_f = 273 - 3 = 270 \ K$ $(q)$.
$b$. $NaCl \rightarrow Na^+ + Cl^-$,$i = 2$. $\Delta T_f = 2 \times 1 = 2 \ K$. $T_f = 273 - 2 = 271 \ K$ $(p)$.
$c$. $K_3[Fe(CN)_6] \rightarrow 3K^+ + [Fe(CN)_6]^{3-}$,$i = 4$. $\Delta T_f = 4 \times 1 = 4 \ K$. $T_f = 273 - 4 = 269 \ K$ $(s)$.
$d$. $Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-}$,$i = 5$. $\Delta T_f = 5 \times 1 = 5 \ K$. $T_f = 273 - 5 = 268 \ K$ $(r)$.
Matching: $a-q, b-p, c-s, d-r$.

(B) Let $A$ be benzene and $B$ be toluene.
Given: $P^o_A = 74.7\,torr$,$P^o_B = 22.3\,torr$,$n_A = 1.5\,mol$,$n_B = 3.5\,mol$.
Total moles $n_{total} = 1.5 + 3.5 = 5.0\,mol$.
Mole fraction of benzene in liquid phase,$x_A = \frac{1.5}{5.0} = 0.3$.
Mole fraction of toluene in liquid phase,$x_B = \frac{3.5}{5.0} = 0.7$.
Total vapour pressure of the solution,$P_{total} = P^o_A x_A + P^o_B x_B = (74.7 \times 0.3) + (22.3 \times 0.7) = 22.41 + 15.61 = 38.02\,torr \approx 38.0\,torr$.
Mole fraction of benzene in vapour phase $(y_A)$ is given by $y_A = \frac{P_A}{P_{total}} = \frac{P^o_A x_A}{P_{total}} = \frac{22.41}{38.02} \approx 0.589$.

(D) For elevation in boiling point: $\Delta T_b = K_b \times m$.
Given $\Delta T_b = 2 \ K$ and $m = 1 \ m$,so $2 = K_b \times 1$,which gives $K_b = 2 \ K \ kg \ mol^{-1}$.
For depression in freezing point: $\Delta T_f = K_f \times m$.
Given $\Delta T_f = 2 \ K$ and $m = 2 \ m$,so $2 = K_f \times 2$,which gives $K_f = 1 \ K \ kg \ mol^{-1}$.
Comparing the two values,$K_b = 2 \ K_f$.

(B) The boiling point of a liquid mixture is the temperature at which its total vapour pressure equals the external pressure.
Given: $P_{total} = 1\, atm = 760\, mm\,Hg$.
$P_A^o = 520\, mm\,Hg$ and $P_B^o = 1000\, mm\,Hg$.
According to Raoult's law: $P_{total} = P_A^o \cdot x_A + P_B^o \cdot x_B$.
Since $x_A + x_B = 1$,we have $x_B = 1 - x_A$.
Substituting the values: $760 = 520 \cdot x_A + 1000 \cdot (1 - x_A)$.
$760 = 520 \cdot x_A + 1000 - 1000 \cdot x_A$.
$480 \cdot x_A = 240$.
$x_A = \frac{240}{480} = 0.5$.
Therefore,the mole percentage of $A$ is $0.5 \times 100 = 50\, mol\,\%$.