Solutions Set-2

Q1. A solution is prepared by dissolving $3.00\ \mathrm{g}$ of glucose ($\mathrm{C_6H_{12}O_6}$, $M=180.16\ \mathrm{g\,mol^{-1}}$) in $90.0\ \mathrm{g}$ of water. Taking $K_f$ for water as $1.86\ \mathrm{K\,kg\,mol^{-1}}$ and using $\Delta T_f = iK_f m$, the depression in freezing point $\Delta T_f$ (in K) is approximately:

Q2. A binary solution is prepared by mixing $40.0\ \mathrm{g}$ of ethanol ($\mathrm{C_2H_5OH}$, $M=46.07\ \mathrm{g\,mol^{-1}}$) with $60.0\ \mathrm{g}$ of water ($M=18.015\ \mathrm{g\,mol^{-1}}$). At the temperature of interest the vapour pressures of pure ethanol and pure water are $P^\circ_{\rm EtOH}=59.0\ \mathrm{kPa}$ and $P^\circ_{\rm H_2O}=31.8\ \mathrm{kPa}$ respectively. Assuming ideal behaviour (Raoult's law), the mole fraction of ethanol in the vapour phase $y_{\rm EtOH}$ is closest to:

Q3. $1.00\ \mathrm{g}$ of sodium sulphate ($\mathrm{Na_2SO_4}$, $M=142.04\ \mathrm{g\,mol^{-1}}$) is dissolved in $100.0\ \mathrm{g}$ of water. If the observed freezing point depression is $\Delta T_f = 0.197\ \mathrm{K}$ and $K_f$ for water is $1.86\ \mathrm{K\,kg\,mol^{-1}}$, the degree of dissociation $\alpha$ of $\mathrm{Na_2SO_4}$ (dissociating as $\mathrm{Na_2SO_4 \rightarrow 2Na^+ + SO_4^{2-}}$, with $i=1+(n-1)\alpha$ and $n=3$) is approximately:

Q4. Equal masses ($1.00\ \mathrm{g}$) of three solutes are dissolved separately in $100.0\ \mathrm{g}$ of water at $25^\circ\mathrm{C}$: (i) glucose ($M=180.16\ \mathrm{g\,mol^{-1}}$, non-electrolyte), (ii) $\mathrm{NaCl}$ ($M=58.44\ \mathrm{g\,mol^{-1}}$, fully dissociates into 2 ions), and (iii) $\mathrm{CaCl_2}$ ($M=110.98\ \mathrm{g\,mol^{-1}}$, fully dissociates into 3 ions). Assuming ideal behaviour and complete dissociation of electrolytes, which solution will have the lowest freezing point?

Q5. A non-volatile solute of mass $9.50\ \mathrm{g}$ is dissolved in $100.0\ \mathrm{g}$ of water and the boiling point of the solution is found to be $0.270\ \mathrm{K}$ higher than that of pure water. Given the ebullioscopic constant $K_b=0.512\ \mathrm{K\,kg\,mol^{-1}}$ and assuming the solute is a non-electrolyte, the molar mass $M$ of the solute (in $\mathrm{g\,mol^{-1}}$) is approximately:

Q6. A and B form an ideal binary solution. At a certain temperature the vapour pressures of pure A and pure B are $P_A^0=40\ \text{kPa}$ and $P_B^0=30\ \text{kPa}$ respectively. A solution is prepared by mixing $46\ \text{g}$ of A (molar mass $=46\ \text{g mol}^{-1}$) and $36\ \text{g}$ of B (molar mass $=18\ \text{g mol}^{-1}$). Assuming ideal behaviour, the total vapour pressure of the solution is closest to:

Q7. $1.50\ \text{g}$ of an electrolyte of formula AB$_2$ (molar mass $=119\ \text{g mol}^{-1}$) is dissolved in $100.0\ \text{g}$ of water. The solution lowers the freezing point by $0.422\ ^\circ\text{C}$. (Take $K_f$ for water $=1.86\ \text{K kg mol}^{-1}$.) Assuming dissociation according to $\mathrm{AB_2 \rightarrow A^{2+} + 2B^-}$ with degree of dissociation $\alpha$, the value of $\alpha$ is closest to:

Q8. Dissolution of $1.00\ \text{g}$ of an electrolyte of formula AB$_2$ in $50.00\ \text{g}$ of benzene causes a depression of freezing point by $1.28\ \text{K}$. (Take $K_f$ of benzene $=5.12\ \text{K kg mol}^{-1}$ and assume complete dissociation, i.e. AB$_2\rightarrow A^{2+}+2B^-$.) The molar mass of the electrolyte is closest to:

Q9. Two volatile liquids A and B form an ideal solution at temperature $T$. The vapour pressures of pure A and B at $T$ are $P_A^0=240\ \text{mmHg}$ and $P_B^0=120\ \text{mmHg}$ respectively. An initial liquid mixture contains $x_A^{(l)}=0.40$ (mole fraction of A). A simple distillation is carried out and $20\%$ of the original amount of liquid (by moles) is vapourized and the vapour is removed as distillate at this temperature. Assuming ideal behaviour and instantaneous equilibrium during vaporization, the mole fraction of A in the remaining liquid is closest to:

Q10. $1.00\ \text{g}$ of $\mathrm{Na_2SO_4}$ (molar mass $=142.04\ \text{g mol}^{-1}$) is dissolved in $250.0\ \text{mL}$ of water at $25^\circ\text{C}$ and the osmotic pressure of the solution is $1.40\ \text{atm}$. Assuming $\mathrm{Na_2SO_4}$ dissociates as $\mathrm{Na_2SO_4 \rightarrow 2Na^+ + SO_4^{2-}}$ with degree of dissociation $\alpha$, calculate $\alpha$ (use $R=0.08206\ \text{L atm mol}^{-1}\text{K}^{-1}$ and $T=298\ \text{K}$):