Solutions MCQ Questions & Answers in Physical Chemistry | Chemistry

$$\eqalign{ & {\text{Total }}V.P.{\text{ of solution}} = P_A^ \circ {x_A} + P_B^ \circ {x_B} \cr & {\text{Given,}}\,P_A^ \circ = 74.7\,torr,P_B^ \circ = 22.3\,torr \cr & {n_{{\text{benzene}}}} = 1.5\,mol,{n_{{\text{toluene}}}} = 3.5\,mol \cr & {n_{{\text{solution}}}} = 1.5 + 3.5 = 5\,mol \cr & {x_A} = \frac{{{n_{{\text{benzene}}}}}}{{{n_{{\text{solution}}}}}} = \frac{{1.5}}{5} \cr & {\text{Total }}V.P.{\text{ of solution}} \cr & {\text{ = }}\left( {\frac{{1.5}}{5} \times 74.7 + \frac{{3.5}}{5} \times 22.3} \right)torr \cr & = \left( {22.4 + 15.6} \right)torr \cr & = 38\,torr \cr & {\text{Mole fraction of benzene in vapour form}} = \frac{{22.4}}{{38}} = 0.589 \cr} $$

$$\Delta {T_f} = i \times {K_f} \times m$$
Where $$m =$$  Molality of the solution ( i.e. number of moles of solute per $$1000 g$$  of the solvent )
$$\eqalign{ & {\text{Here}}\,\,m = \frac{{0.1}}{{329}} \times 100 \cr & {\text{Thus}}\,\,\Delta {T_f} = 4 \times 1.86 \times \frac{{0.1 \times 100}}{{329}} = 2.3 \times {10^{ - 2}} \cr & {\text{Thus}}\,\,{T_f} = 0 - 2.3 \times {10^{ - 2}} = -2.3 \times {10^{ - 2}}{\,^ \circ }C \cr} $$

$$\eqalign{ & {\text{Van't}}\,\,{\text{Hoff equation is}} \cr & \pi V = inRT \cr & {\text{For depression in freezing point}}. \cr & \Delta {T_f} = i \times {K_f} \times m \cr & {\text{For elevation in boiling point}}{\text{.}} \cr & \Delta {T_b} = i \times {K_b} \times m \cr & {\text{For lowering of vapour pressure,}} \cr & \frac{{P_{solvent}^ \circ - P_{solution}^ \circ }}{{P_{solvent}^ \circ }} = i\left( {\frac{n}{{N + n}}} \right) \cr} $$

$$\eqalign{ & {\text{According to depression of freezing point,}} \cr & \Delta {T_f} = {k_f} \times {\text{molality of solution}} \cr & {\text{According to elevation of boiling point,}} \cr & \Delta {T_b} = {k_b} \times {\text{molality of solution}} \cr & {\text{or}}\,\,\frac{{\Delta {T_f}}}{{\Delta {T_b}}} = \frac{{{k_f}}}{{{k_b}}} \cr & {\text{Given that}} \cr & \Delta {T_b} = {T_2} - {T_1} \cr & \,\,\,\,\,\,\,\,\,\,\, = 100.18 - 100 \cr & \,\,\,\,\,\,\,\,\,\,\, = 0.18 \cr & {K_f}\,{\text{for water}} = 1.86\,K\,kg\,mo{l^{ - 1}} \cr & {k_{b\,}}\,{\text{for water}} = 0.512\,K\,kg\,mo{l^{ - 1}} \cr & \therefore \,\,\frac{{\Delta {T_f}}}{{0.18}} = \frac{{1.86}}{{0.512}} \cr & \,\,\,\,\,\,\,\,\Delta {T_f} = \frac{{1.86 \times 0.18}}{{0.512}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.6539 \approx 0.654 \cr & \,\,\,\,\Delta {T_f} = {T_1} - {T_2} \cr & 0.654\,\, = {0^ \circ }C - {T_2} \cr & \therefore \,\,\,{T_2} = - {0.654^ \circ }C \cr} $$
( $${T_2} \to $$   freezing point of aqueous urea solution )

When blood cells are placed in a solution of similar concentration as that of blood, then they neither swell nor shrink it means the concentration of solution is same as that of inside the blood cells, i.e. they are isotonic to each other.

$$\eqalign{ & \Delta {T_f}\left( {{\text{normal}}} \right) = {K_f}m \cr & = 1.86 \times 0.01 \cr & = 0.0186\,; \cr & i = \frac{{\Delta {T_{f\left( {obs} \right)}}}}{{\Delta {T_{f\left( {nor} \right)}}}} \cr & = \frac{{0.0205}}{{0.0186}} \cr & = 1.10 \cr & = 1 + \alpha \,; \cr & \alpha = 0.1 \cr & {K_a} = \frac{{C{\alpha ^2}}}{{1 - \alpha }} \cr & = \frac{{0.01 \times {{0.1}^2}}}{{1 - 0.1}} \cr & = \frac{1}{9} \times {10^{ - 3}}; \cr & {K_b} = \frac{{{K_w}}}{{{K_a}}} \cr & = 1.0 \times {10^{ - 14}} \times 9 \times {10^3} \cr & = 9 \times {10^{ - 11}} \cr} $$

$$\eqalign{ & {P_{total}} = P_A^ \circ \times {X_A} + P_B^ \circ \times {X_B} \cr & = 80.0 \times 0.4 + 120.0 \times 0.6 = 104\,mm\,Hg \cr} $$
The observed $${P_{total}}$$  is $$100\,mm\,Hg$$   which is less than $$104\,mm\,Hg.$$   Hence the solution shows negative deviation.

Isotonic solutions are the solutions having same osmotic pressure.
Osmotic pressure of $$5\% $$  cane sugar solution
$$\eqalign{ & \left( {{\pi _1}} \right) = C \times R \times T \cr & \,\,\,\,\,\,\,\,\,\,\, = \frac{{50g/L}}{{342}} \times 0.0821 \times T \cr} $$
Osmotic pressure of $$1\% $$  solution of substance
$$\eqalign{ & X\left( {\pi 2} \right) = \frac{{10g/L}}{M} \times 0.0821 \times T \cr & {\text{Both are isotonic}} \cr & {\text{So,}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\pi _1} = {\pi _2} \cr & {\text{or}}\,\,\frac{{50}}{{342}} \times 0.0821 \times T = \frac{{10}}{M} \times 0.0821 \times T \cr} $$
∴ $$M$$ ( molecular weight of $$X$$ ) $$ = \frac{{342}}{5} = 68.4$$

$$\eqalign{ & {\text{Osmotic pressure}}\,\left( \pi \right) = CRT \cr & \left( \pi \right) = \frac{{wt \times 1000}}{{{\text{Molecular mass}} \times V}}RT \cr & = 2.57 \times {10^{ - 3}} \cr & = \frac{{1.26 \times 1000}}{{{\text{Mol}}{\text{.mass}} \times 200}} \times 0.083 \times 300 \cr & {\text{Molecular mass}} = 61038\,g \cr} $$

0.01 molal $$NaCl$$  solution have minimum freezing point because its molality is more and it gives two ions after dissociation of one molecule.