Solutions Set-9

Q1. 3.00 g of a non-electrolytic solute (molar mass = $150.0\ \mathrm{g\,mol^{-1}}$) is dissolved in $90.0\ \mathrm{g}$ of benzene. The molal freezing point depression constant for benzene is $K_f=5.12\ \mathrm{K\,kg\,mol^{-1}}$ and the freezing point of pure benzene is $5.50\,^\circ\mathrm{C}$. Using $\Delta T_f = K_f\times m$ where $m$ is the molality, the freezing point of the solution is closest to:

Q2. A binary solution of volatile liquids X and Y obeys Raoult's law. At $25^\circ\mathrm{C}$ the vapour pressures of pure X and Y are $P_X^0=400\ \mathrm{mmHg}$ and $P_Y^0=100\ \mathrm{mmHg}$ respectively. The vapour above the solution contains $60.0\%$ X by mole (i.e. $y_X=0.60$). Using $y_X=\dfrac{x_X P_X^0}{x_X P_X^0+(1-x_X)P_Y^0}$, the mole fraction $x_X$ of X in the liquid phase is closest to:

Q3. A sample of an ionic compound AB$_2$ is dissolved so that $0.10\ \mathrm{mol}$ of AB$_2$ is present in $200.0\ \mathrm{g}$ of water. The boiling point elevation of the solution is observed to be $\Delta T_b=0.5632\ \mathrm{K}$. The ebullioscopic constant for water is $K_b=0.512\ \mathrm{K\,kg\,mol^{-1}}$. Assuming partial dissociation with van 't Hoff factor $i=1+2\alpha$ (where $\alpha$ is the degree of dissociation), the value of $\alpha$ is closest to: (use $\Delta T_b=iK_b m$ and $m=\dfrac{\text{moles solute}}{\text{kg solvent}}$)

Q4. Consider the following two statements about a non-volatile solute dissolved in a solvent:

Statement A: The chemical potential of the solvent decreases on addition of a non-volatile solute, i.e. $\mu_{\text{solvent}}^{\text{(solution)}}<\mu_{\text{solvent}}^{\text{(pure)}}$, which leads to lowering of the solvent vapour pressure.

Statement R: This decrease in chemical potential (and the resulting vapour pressure lowering) is primarily caused by an enthalpy change on mixing (heat of mixing) that stabilizes solvent molecules in the solution.

Choose the correct option:

Q5. $0.600\ \mathrm{g}$ of a compound X (molar mass $M=60.0\ \mathrm{g\,mol^{-1}}$) is dissolved in $100.0\ \mathrm{g}$ of benzene. The observed freezing point depression is $\Delta T_f=0.384\ \mathrm{K}$. The cryoscopic constant for benzene is $K_f=5.12\ \mathrm{K\,kg\,mol^{-1}}$. If X partially dimerizes in benzene according to

with degree of association $\alpha$ (fraction of monomers that dimerize), the value of $\alpha$ is closest to:

Q6. 0.50 g of a non-volatile, non-electrolyte solute is dissolved in enough water to make $200\ \mathrm{mL}$ of solution at $25^\circ\mathrm{C}$ ($T=298\ \mathrm K$). The osmotic pressure of the solution is $0.306\ \mathrm{atm}$. Using $\pi=cRT$ and $R=0.08206\ \mathrm{L\,atm\,mol^{-1}K^{-1}}$, the molar mass of the solute is closest to:

Q7. 1.80 g of an ionic compound of the type $X\mathrm{Cl}_2$ is dissolved in $100\ \mathrm{g}$ of water and the freezing point is lowered by $1.86\ \mathrm{K}$. For water $K_f=1.86\ \mathrm{K\,kg\,mol^{-1}}$. Assuming the salt dissociates as $X^{2+}+2\mathrm{Cl}^-$ with degree of dissociation $\alpha=0.50$, use $\Delta T_f=K_f m i$ and $i=1+2\alpha$ to find the molar mass of $X\mathrm{Cl}_2$:

Q8. Assertion (A): For a binary solution showing negative deviation from Raoult's law, the total vapour pressure of the solution is greater than that predicted by Raoult's law.
Reason (R): Negative deviation occurs because A–B interactions are stronger than A–A and B–B interactions, so the enthalpy of mixing is negative ($\Delta H_{\mathrm{mix}}<0$), stabilizing the liquid and lowering the vapour pressure.

Q9. Two solutes, $2.00\ \mathrm g$ of urea (molar mass $60.0\ \mathrm{g\,mol^{-1}}$) and $2.00\ \mathrm g$ of NaCl (molar mass $58.44\ \mathrm{g\,mol^{-1}}$), are dissolved in $100.0\ \mathrm g$ of water. If NaCl dissociates to the extent $\alpha=0.90$, calculate the freezing point depression $\Delta T_f$. Use $K_f=1.86\ \mathrm{K\,kg\,mol^{-1}}$ and the relation $\Delta T_f=K_f\sum m_i i_i$ (molality $m$ in mol kg$^{-1}$; for NaCl $i=1+\alpha$):

Q10. A solution at $298\ \mathrm K$ contains $30.0\%$ ethanol (C$_2$H$_5$OH, $M=46.07\ \mathrm{g\,mol^{-1}}$) by mass and $70.0\%$ water by mass. Vapour pressures of pure ethanol and pure water at $298\ \mathrm K$ are $P^\circ_{\mathrm{eth}}=59.0\ \mathrm{mmHg}$ and $P^\circ_{\mathrm{water}}=23.8\ \mathrm{mmHg}$ respectively. Assuming an ideal solution (Raoult's law), the mole fraction of ethanol in the vapour phase above the solution, $y_{\mathrm{ethanol}}$, is closest to:

Q11. 0.50 g of a non-volatile, non-electrolyte solute is dissolved in 25.0 g of benzene and the freezing point is lowered by $0.64^\circ\text{C}$. Given $K_f(\text{benzene})=5.12\ \text{K kg mol}^{-1}$, the molar mass of the solute is:

Q12. 10.0 g of benzene ($M=78.11\ \mathrm{g\,mol^{-1}}$) and 20.0 g of toluene ($M=92.14\ \mathrm{g\,mol^{-1}}$) are mixed at a temperature where the vapour pressures of pure benzene and pure toluene are $100\ \mathrm{mmHg}$ and $50\ \mathrm{mmHg}$ respectively. Assuming an ideal solution, the mole fraction of benzene in the vapour phase is approximately:

Q13. A sample of an electrolyte of unknown molar mass (assumed to dissociate completely into two ions) dissolves when $0.146\ \mathrm{g}$ is added to $100.0\ \mathrm{g}$ of water. The osmotic pressure of the resulting solution at $25^\circ\mathrm{C}$ ($298\ \mathrm{K}$) is measured to be $0.600\ \mathrm{atm}$. (Assume the volume of the solution ≈ $100.0\ \mathrm{mL}$.) The molar mass of the electrolyte is closest to:

Q14. Assertion (A): In dilute aqueous solutions of strong electrolytes, experimentally measured colligative effects are often smaller than the values calculated assuming complete dissociation.
Reason (R): This discrepancy arises because ions in solution attract each other and form ion pairs (or have inter-ionic interactions), effectively reducing the number of free particles contributing to colligative properties.

Q15. A weak electrolyte AB dissociates as $\mathrm{AB} \rightleftharpoons \mathrm{A}^{+}+\mathrm{B}^{-}$. A $0.010\ \mathrm{molal}$ aqueous solution of AB shows colligative behaviour corresponding to an apparent van't Hoff factor $i=1.80$. Estimate the degree of dissociation $\alpha$ of AB at this concentration and state qualitatively how $\alpha$ will change if the concentration is increased to $0.040\ \mathrm{molal}$.