$$\eqalign{ & {\text{Given}}\,\,{\pi _{ob}} = 10.8\,atm \cr & {\pi _{nor}} = CRT = 0.10 \times 0.0821 \times 298 = 2.446 \cr & {\text{now experimental value of }}(i) \cr & = \frac{{{\text{Observed osmotic pressure}}}}{{{\text{Normal osmotic pressure}}}} \cr & = \frac{{10.8}}{{2.446}} \cr & = 4.42 \cr} $$

$$\eqalign{ & {\text{From the molarity equation}}{\text{.}} \cr & {M_1}{V_1} + {M_2}{V_2} = MV \cr & M = \frac{{{M_1}{V_1} + {M_2}{V_2}}}{V}{\text{where}}\,\,V = {V_1} + {V_2} \cr & M = \frac{{480 \times 1.5 + 520 \times 1.2}}{{480 + 520}} \cr & = 1.344\,M \cr} $$

For a dilute solution, the depression in freezing point $$\left( {\Delta {T_f}} \right)$$  is directly proportional to molality $$(m)$$ of the solution.
$$\Delta {T_f} \propto m\,\,\,\,or\,\,\,\Delta {T_f} = {K_f}m$$
Where, $${{K_f}}$$ is called molal depression constant or freezing point depression constant or cryoscopic constant. The value of $${{K_f}}$$ depends only on nature of the solvent and independent of composition of solute particles, i.e. does not depend on the concentration of solution.

The solutions (liquid mixture) which boils at constant temperature and can distil as such without any change in composition are called azeotropes.
Solution of $$HN{O_3}$$  and $${H_2}O$$  will form maximum boiling point azeotrope. Maximum boiling azeotropes show negative deviation from Raoult’s law.
$$\matrix{ {} & {{\rm{Composition}}\left( \% \right)} & {{\rm{Boiling Point }}} \cr {HN{O_3}} & {68.0} & {359\,K} \cr {{H_2}O} & {32.0} & {373\,K} \cr } $$
Boiling point of the azeotrope of these two solutions is $$393.5\,K.$$

$$\eqalign{ & HX\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \rightleftharpoons \,\,\,{H^ + } + {X^ - } \cr & {\text{Weak acid}} \cr & 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\left( {{\text{initially}}} \right) \cr & 1 - \alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\alpha \,\,\,\,\left( {{\text{at equilibrium}}} \right) \cr & {\text{Total}} = 1 - \alpha + \alpha + \alpha \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 + \alpha \cr & \,\,\,\,\,\,\,\,\,\,i = \frac{{1 + \alpha }}{1} \cr & \,\,\,\,\,\,\,\,\alpha = 20\% \,{\text{dissociation}} \cr & {\text{i}}{\text{.e}}{\text{.}}\,\,\alpha = 0.2 \cr & i = 1 + \alpha \cr & \,\,\, = 1 + 0.2 \cr & \,\,\, = 1.2 \cr & \Delta {T_f} = i \times {K_f} \times m \cr & \,\,\,\,\,\,\,\,\,\,\, = 1.2 \times 1.86\,K\,kg\,mo{l^{ - 1}} \times 0.5 \cr & \,\,\,\,\,\,\,\,\,\,\, = 1.12\,K \cr} $$

$${C_6}{H_{12}}{O_6}$$   is a non-electrolyte hence furnishes minimum number of particles and will have maximum freezing point.
$$\Delta {T_f} = i{K_f}m\,\,\,\,{\text{or}}\,\,\,\Delta {T_f} \propto i\,\,\,\,{\text{and}}\,\,\,\Delta {{\text{T}}_f} = T_f^ \circ - {T_f}$$

$$\eqalign{ & {\text{From Molarity equation}} \cr & {M_1}{V_1} + {M_2}{V_2} = M \times V \cr & M = \frac{{{M_1}{V_1} + {M_2}{V_2}}}{V}\,{\text{where}}\,{\text{V = total}}\,{\text{volume}} \cr & {\text{ = }}\frac{{750 \times 0.5 + 250 \times 2}}{{1000}} = 0.875M \cr} $$

$$\eqalign{ & {\text{Normality}} = 1.5\,N \cr & {\text{Equivalent weight of}}\,{H_2}{O_2} = 17 \cr & {\text{So, strength of the solutions,}} \cr & S = E \times N \cr & \,\,\,\,\, = 17 \times 1.5 = 25.5 \cr & \,\,\,\,\,\,\,\,\,\,2{H_2}{O_2} \to 2{H_2}O + {O_2} \cr & \,\,\,\,\, = 2 \times 34 \cr & \,\,\,\,\, = 68\,g \cr} $$
$$\because \,\,68\,g$$   of $${H_2}{O_2}$$  produce $${O_2}$$  at $$NTP = 22.4\,L$$
$$\therefore 25.5\,g$$   of $${H_2}{O_2}$$  will produce
$$\eqalign{ & = \frac{{22.4}}{{68}} \times 25.5 \cr & = 8.4\,L\,\,{\text{of}}\,{O_2} \cr} $$

$$\eqalign{ & {\text{Using the relation,}} \cr & M = \frac{{w.R.T}}{{\pi V}}, \cr & {\text{we get}}\,M = \frac{{6 \times 0.080 \times 300}}{{2 \times {{10}^{ - 3}} \times 0.2}} \cr & = 3.6 \times {10^5} \cr} $$

$$\eqalign{ & {\text{The given data are}} \cr & {P_{water}} = 17.0\,torr; \cr & {P_{total}}\left( {4\,mole\,\% \,{\text{solution}}} \right) \cr & = {P_{N{H_3}}} + {P_{water}} = 50.0\,torr \cr & {X_{N{H_3}}} = 0.04\,\,{\text{and}}\,\,{X_{water}} = 0.96 \cr & {\text{Now according to Raoult's}}\,{\text{law;}} \cr & {P_{water}} = {X_{water}}\,P_{water}^ \circ \cr & = 0.96 \times 17.0\,torr = 16.32\,torr \cr & {\text{Now Henry's}}\,{\text{law constant for ammonia is}} \cr & {K_H}\left( {N{H_3}} \right) = \frac{{{P_{N{H_3}}}}}{{{X_{N{H_3}}}}} = \frac{{33.68\,torr}}{{0.04}} = 842\,torr \cr & {\text{Hence, for}}\,\,5\,mole\,\% \,\,{\text{solution, we have}} \cr & {P_{N{H_3}}} = {K_H}\left( {N{H_3}} \right){X_{N{H_3}}} \cr & = \left( {842\,torr} \right)\left( {0.05} \right) = 16.15\,torr \cr & {\text{Thus,}}\,{P_{total}}\left( {5\,mole\,\% \,{\text{solution}}} \right) \cr & = {P_{N{H_3}}} + {P_{water}} \cr & = 42.1 + 16.15 \cr & = 58.25\,torr \cr} $$