Solutions MCQ Questions & Answers in Physical Chemistry | Chemistry

5.2 molal solution means 5.2 moles of methyl alcohol in 1000 gm water or in $$\frac{{1000}}{{18}}$$ mole of water.
∴ mole fraction of methyl alcohol
$$\eqalign{ & = \frac{{{\text{moles of methyl alcohol}}}}{{{\text{moles of methyl alcohol + moles of water}}}} \cr & = \frac{{5.2}}{{5.2 + \frac{{1000}}{{18}}}} \cr & = 0.086 \cr} $$

No. of moles of urea $$ = \frac{{10}}{{60}} = \frac{1}{6}$$
Weight of solute, $$X = 6\,g$$
No. of moles of $$X = \frac{6}{M}$$
For isotonic solutions, $${n_1} = {n_2}$$   or $$\frac{1}{6} = \frac{6}{M}$$   or $$M = 36\,g\,mo{l^{ - 1}}$$

$$\Delta {T_f}{\text{(normal)}} = {K_f}m = 1.86 \times 0.5 = {0.93^ \circ }$$         Assuming 100% ionization, $$i = n = No.$$  of $$ions$$  per molecule
$$\eqalign{ & = \frac{{{\text{Observed}}\,\Delta {T_f}}}{{{\text{Normal}}\,\Delta {T_f}}} \cr & = \frac{{3.72}}{{0.93}} \cr & = 4 \cr & {K_3}\left[ {Fe{{\left( {CN} \right)}_6}} \right] \to 3{K^ + } + {\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 - }} \cr & {\text{no}}{\text{. of }}ions = 4 \cr} $$

$$10\,g$$  per $$d{m^3}$$  of urea is isotonic with  $$5\% $$  solution of a non-volatile solute. Hence, between these solutions osmosis is not possible, so their molar concentrations are equal to each other.
Thus, molar concentration of urea solution
$$\eqalign{ & = \frac{{10\,g/\,d{m^3}}}{{{\text{Molecular weight of urea}}}} \cr & = \frac{{10}}{{60}}M \cr & = \frac{1}{6}\,M \cr} $$
Molar concentration of  $$5\% $$  non-volatile solute
$$\eqalign{ & = \frac{{50\,g/\,d{m^3}}}{{{\text{Molecular weight of non - volatile solute}}}} \cr & = \frac{{50}}{m}M \cr} $$
Both solutions are isotonic to each other, therefore
$$\eqalign{ & \frac{1}{6} = \frac{{50}}{m} \cr & {\text{or}}\,\,m = 50 \times 6 \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 300\,g\,mo{l^{ - 1}} \cr} $$

Reverse osmosis will occur.

Reverse osmosis is used for desalination of sea water.

$$\eqalign{ & {\text{No}}{\text{. of moles of }}NaCl \cr & = \frac{{M \times V}}{{1000}} \cr & = \frac{{0.25 \times 100}}{{1000}} \cr & = 0.025 \cr & NaCl \to N{a^ + } + C{l^ - } \cr & {\text{No}}{\text{. of moles of}}\,\,N{a^ + }\,ions = 0.025 \cr & {\text{No}}{\text{. of}}\,\,N{a^ + }\,ions = 0.025 \times 6.023 \times {10^{23}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1.505 \times {10^{22}} \cr} $$

For ideal solution vapour pressure of solution
$$\eqalign{ & P_A^ \circ {X_A} + P_B^ \circ {X_B} \cr & = 80 \times \frac{2}{5} + 100 \times \frac{3}{5} \cr & = 92\,torr \cr} $$
Since observed vapour pressure of solution < ideal vapour pressure, the solution shows negative deviation

$$\eqalign{ & {\text{No}}{\text{. of moles of polymer}} = \frac{1}{{150,000}} \cr & \pi = CRT = \frac{n}{V}RT; \cr & \pi = \frac{1}{{150,000}} \times \frac{{8.314 \times {{10}^3} \times 310}}{{0.5}} \cr & \,\,\,\, = 34.36\,Pa\,\,\,\left( {R = 8.314 \times {{10}^3}\,Pa\,\,L\,\,{K^{ - 1}}\,\,mo{l^{ - 1}}} \right) \cr} $$

$$\eqalign{ & P_M^o = 450\,mmHg,P_N^o = 700\,mmHg \cr & {P_M} = P_M^o{x_M} = {y_M}{P_T} \cr & \Rightarrow P_M^o = \frac{{{y_M}}}{{{x_M}}}\left( {{P_T}} \right) \cr & {\text{Similarly,}}\,P_N^o = \frac{{{y_N}}}{{{x_N}}}\left( {{P_T}} \right) \cr & {\text{Given,}}\,P_M^o < P_N^o \cr & \Rightarrow \frac{{{y_M}}}{{{x_M}}} < \frac{{{y_N}}}{{{x_N}}} \cr & \Rightarrow \frac{{{y_M}}}{{{y_N}}} < \frac{{{x_M}}}{{{x_N}}} \cr} $$