Solutions Set-5

Q1. 0.50 g of an unknown non-electrolyte solute is dissolved in 100.0 g of benzene. The freezing point of pure benzene is $5.53^\circ\text{C}$ and the solution freezes at $4.87^\circ\text{C}$. Given $K_f=5.12\ \text{K kg mol}^{-1}$ for benzene, the molar mass of the solute is:

Q2. $2.77\ \text{g}$ of $\text{CaCl}_2$ ($M=110.98\ \text{g mol}^{-1}$) is dissolved in $100.0\ \text{g}$ of water and the freezing point is depressed by $0.93\ \text{K}$. Given $K_f=1.86\ \text{K kg mol}^{-1}$ for water, the degree of dissociation $\alpha$ of $\text{CaCl}_2$ is approximately:

Q3. At a fixed temperature the vapor pressures of pure liquids A and B are $p_A^\circ=300\ \text{mmHg}$ and $p_B^\circ=100\ \text{mmHg}$ respectively. An ideal solution of A and B shows total vapor pressure $P_{\text{total}}=220\ \text{mmHg}$. Assuming Raoult's law, the mole fraction of A in the liquid $x_A$ and the mole fraction of A in the vapor $y_A$ are:

Q4. Statement I: For an ideal binary solution of volatile liquids A and B at a given temperature, if $p_A^\circ>p_B^\circ$, then for any composition $0<x_A<1$ the mole fraction of A in the vapor phase $y_A$ is greater than its mole fraction in the liquid phase $x_A$.
Statement II: This is because $y_A=\dfrac{x_A p_A^\circ}{x_A p_A^\circ + x_B p_B^\circ}$ and $p_A^\circ>p_B^\circ$ mathematically implies $y_A>x_A$ for $0<x_A<1$.

Q5. $3.00\ \text{g}$ of a carboxylic acid (monomer molar mass $M=60.0\ \text{g mol}^{-1}$) is dissolved in $100.0\ \text{g}$ of benzene. The observed boiling point elevation is $\Delta T_b=0.88\ \text{K}$. Given $K_b=2.53\ \text{K kg mol}^{-1}$ for benzene and that the acid dimerizes in solution via $2X\rightleftharpoons X_2$, the degree of association $\alpha$ is approximately:

Q6. A non-volatile solute is dissolved in benzene at 298 K. The vapour pressure of pure benzene at 298 K is $P^\circ_{\mathrm{benzene}} = 100\ \mathrm{mmHg}$ and the mole fraction of benzene in the solution is $x_{\mathrm{benzene}} = 0.90$. Using Raoult's law $P = x_{\mathrm{benzene}}P^\circ_{\mathrm{benzene}}$, the vapour pressure of benzene above the solution is:

Q7. A sample of mass $1.00\ \text{g}$ of a non-electrolyte is dissolved in $100.0\ \text{g}$ of water. The freezing point of the solution is lowered by $\Delta T_f = 0.62\ \text{K}$. Given $K_f(\text{water}) = 1.86\ \text{K\,kg\,mol}^{-1}$, calculate the molar mass of the solute. (Use $\Delta T_f = K_f m$ and $m=\dfrac{\text{moles solute}}{\text{kg solvent}}$.)

Q8. One mole of an ionic compound of formula AB$_2$ is dissolved in $1.00\ \text{kg}$ of water. The observed freezing point depression is $\Delta T_f = 3.72\ \text{K}$. Given $K_f(\text{water})=1.86\ \text{K\,kg\,mol}^{-1}$ and assuming partial dissociation $AB_2 \rightleftharpoons A^{2+}+2B^{-}$ with degree of dissociation $\alpha$, use $\Delta T_f = K_f m i$ and $i = 1 + 2\alpha$ to find $\alpha$.

Q9. Two volatile liquids A and B form an ideal solution at a given temperature. Their vapour pressures are $P_A^\circ = 400\ \mathrm{mmHg}$ and $P_B^\circ = 200\ \mathrm{mmHg}$. A liquid initially contains $1.00\ \mathrm{mol}$ A and $1.00\ \mathrm{mol}$ B and is in equilibrium with its vapour. (i) Calculate the mole fraction of A in the vapour, $y_A$. (ii) If $0.50\ \mathrm{mol}$ of the vapour (at the equilibrium composition) is removed and condensed (and separated), what is the mole fraction of A remaining in the liquid? (Give the final liquid mole fraction.)

Q10. A solute A associates in a non-polar solvent according to $2A \rightleftharpoons A_2$. For an initial analytical concentration $c_0 = 0.010\ \mathrm{mol\,L^{-1}}$ the association equilibrium constant (in concentration units) is given as

If $\alpha$ is the fraction of monomeric A that has dimerized, with $[A]=c_0(1-\alpha)$ and $[A_2]=\dfrac{c_0\alpha}{2}$, find $\alpha$.

Q11. A student wants to lower the freezing point of $500.0\ \mathrm{g}$ of water by $2.00\ ^\circ\mathrm{C}$. How many grams of ethylene glycol (C$_2$H$_6$O$_2$, $M=62.0\ \mathrm{g\,mol^{-1}}$) must be added? ($K_f$ for water $=1.86\ \mathrm{K\,kg\,mol^{-1}}$). Use $\Delta T_f=K_f m$ and treat ethylene glycol as a non‑electrolyte.

Q12. At $60^\circ\mathrm{C}$ the vapour pressures of pure liquids A and B are $P^\circ_A=400\ \mathrm{mmHg}$ and $P^\circ_B=200\ \mathrm{mmHg}$. An ideal solution contains $1.00$ mol A and $x$ mol B and its total vapour pressure is $300\ \mathrm{mmHg}$. Using Raoult's law $P_{\text{tot}}=x_AP^\circ_A+x_BP^\circ_B$, determine the mole fraction of A in the vapour phase $y_A$.

Q13. $6.00\ \mathrm{g}$ of an organic compound ($M=60.0\ \mathrm{g\,mol^{-1}}$) is dissolved in $500.0\ \mathrm{g}$ of water. The observed freezing point depression is $\Delta T_f=0.298\ \mathrm{K}$. ($K_f$ for water $=1.86\ \mathrm{K\,kg\,mol^{-1}}$). Assuming the solute associates to form dimers in solution, calculate the degree of association $\alpha$ (fraction of monomers dimerised). Use $\Delta T_f=iK_f m$ and for dimerisation $i=1-\tfrac{\alpha}{2}$.

Q14. Assertion (A): A binary mixture that shows positive deviation from Raoult's law can boil at a temperature lower than either pure component (i.e., form a minimum‑boiling azeotrope).
Reason (R): Positive deviation occurs because A–B interactions are weaker than A–A and B–B interactions, which raises the vapour pressure of the mixture above the ideal value.

Q15. A $0.500\ \mathrm{g}$ sample of a polymer is dissolved to make $250.0\ \mathrm{mL}$ of solution. The osmotic pressure at $300.0\ \mathrm{K}$ is $0.03730\ \mathrm{atm}$. Using $\pi=\dfrac{n}{V}RT$ for a dilute solution (take $R=0.08206\ \mathrm{L\,atm\,mol^{-1}\,K^{-1}}$), calculate the molar mass of the polymer (in $\mathrm{g\,mol^{-1}}$).