$$\eqalign{ & {P_{{\text{Total}}}} = {P^ \circ }_A{x_A} + {P^ \circ }_B{x_B} \cr & = {P^ \circ }_{{\text{Heptane}}}\,{x_{{\text{Heptane}}}} + {P^ \circ }_{{\text{Octane}}}\,{x_{{\text{Octane}}}} \cr & = 105 \times \frac{{\frac{{25}}{{100}}}}{{\frac{{25}}{{100}} + \frac{{35}}{{114}}}} + 45 \times \frac{{\frac{{35}}{{114}}}}{{\frac{{25}}{{100}} \times \frac{{35}}{{114}}}} \cr & = 105 \times \frac{{0.25}}{{0.25 + 0.3}} + 45 \times \frac{{0.3}}{{0.25 + 0.3}} \cr & = \frac{{105 \times 0.25}}{{0.55}} + \frac{{45 \times 0.3}}{{0.55}} \cr & = \frac{{26.25 + 13.5}}{{0.55}} \cr & = 72\,kPa \cr} $$

$$\eqalign{ & {V_{{\text{dissolved}}}} = K{P_{{H_2}}} \cr & {P_{{H_2}}}\,{\text{in the mixture}} = {X_{{H_2}}}P = \frac{{80 \times 1425}}{{100}} = 1140 \cr & torr = \frac{{1140}}{{760}} = 1.5\,atm \cr & {\text{Hence,}}\,\,{V_{{\text{dissolved}}}}\,{\text{in}}\,\,2L\,{\text{of water}} = 2 \times K \times {P_{{H_2}}} \cr & = 2 \times 18 \times 1.5\, = 54\,mL \cr} $$

\[\underset{\begin{smallmatrix} 1 \\ 1-\alpha \end{smallmatrix}}{\mathop{N{{a}_{2}}S{{O}_{4}}}}\,\rightleftharpoons \underset{\begin{smallmatrix} 0 \\ 2\alpha \end{smallmatrix}}{\mathop{2N{{a}^{+}}}}\,+\underset{\begin{smallmatrix} 0 \\ \alpha \end{smallmatrix}}{\mathop{SO_{4}^{2-}}}\,\]
$${\text{van't Hoff factor}}\left( {\text{i}} \right) = \frac{{1 - \alpha + 2\alpha + \alpha }}{1} = 1 + 2\alpha $$

$$\eqalign{ & \Delta {T_f} = i{K_f} \times m \cr & \,\,\,\,\,\,\,\,\,\,\, = 2 \times 1.86 \times 0.5 \cr & \,\,\,\,\,\,\,\,\,\,\, = {1.86^ \circ }C \cr & {T_f} = T_f^ \circ - \Delta {T_f} \cr & \,\,\,\,\,\,\, = 0 - 1.86 \cr & \,\,\,\,\,\,\, = - {1.86^ \circ }C \cr} $$

Depression in freezing point $$ \propto $$ number of particles.
In colligative properties ions behave like particles.
$$A{l_2}{\left( {S{O_4}} \right)_3}$$   provides five ions on ionisation as
$$\eqalign{ & A{l_2}{\left( {S{O_4}} \right)_3} \to 2\,A{l^{3 + }} + 3SO_4^{2 - } \cr & KI \rightleftharpoons {K^ + } + {I^ - } \cr} $$
while $$KI$$  provides two ions and $${C_5}{H_{10}}{O_5}$$   and $${C_{12}}{H_{22}}{O_{11}}$$   are not ionised, so they have single particle
Hence, lowest freezing point is possible for $$A{l_2}{\left( {S{O_4}} \right)_3}.$$

$$\eqalign{ & \pi V = \frac{w}{m}RT \cr & \therefore \,\,6 \times {10^{ - 4}} \times 1 = \frac{4}{m} \times 0.0821 \times 300; \cr & m = 1.64 \times {10^5}. \cr} $$

For this solution intermolecular interactions between $$n$$ - heptane and ethanol are weaker than $$n$$ - heptane - $$n$$ - heptane & ethanol-ethanol interactions hence the solution of $$n$$ - heptane and ethanol is non-ideal and shows positive deviation from Raoult’s law.

$$\eqalign{ & P_A^o = ?\,,\,{\text{Given}}\,P_B^o = 200mm,\,\,{X_A} = 0.6, \cr & {x_B} = 1 - 0.6 = 0.4,\,P = 290 \cr & P = {P_A} + {P_B} = P_A^o{x_A} + P_B^o{x_B} \cr & \Rightarrow 290 = P_A^o\, \times \,0.6 + 200 \times 0.4 \cr & \therefore \,\,P_A^o = 350\,mm \cr} $$

At 1 atmospheric pressure the boiling point of mixture is $${80^ \circ }C.$$
boiling point the vapour pressure of mixture, $${P_T} = 1$$
atmosphere $$ = 760\,mm\,Hg.$$
Using the relation,
$$\eqalign{ & {P_T} = P_A^o{X_A} + P_B^o{X_B}\,,\,{\text{we}}\,\,{\text{get}} \cr & {P_T} = 520{X_A} + 1000\left( {1 - {X_A}} \right) \cr & \left\{ {\because \,P_A^o = 520\,mm\,Hg,\,\,P_B^o = 1000\,mm\,Hg,\,\,{X_A} + {X_B}\left. { = 1} \right\}} \right. \cr & {\text{or}}\,\,760 = 520{X_A} + 1000 - 1000{X_A}\,\,{\text{or}}\,\,480{X_A} = 240 \cr & {\text{or}}\,\,{X_A} = \frac{{240}}{{480}} = \frac{1}{2}\,\,{\text{or}}\,50\,{\text{mol}}{\text{.}}\,{\text{percent}} \cr & {\text{i}}{\text{.e}}{\text{.}}\,{\text{,}}\,{\text{The correct answer is (D)}} \cr} $$

In hypotonic solution, the water is drawn in and the grape swells while in hypertonic solution the water is drawn out and the grape shrinks.