$$\eqalign{ & \Delta {T_b} = {K_b} \times m \cr & {\text{or}}\,\,\,{K_b} = \frac{{\Delta {T_b}}}{m} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{{\frac{1}{{50}} \times \frac{{1000}}{{50}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{5}{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 2.5 \cr} $$

$$\eqalign{ & {\text{From Raoult's law of partial pressure,}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\frac{{p_A^ \circ - {p_S}}}{{{p_S}}} = \frac{{{n_B}}}{{{n_A}}} \cr & \Rightarrow \frac{{760 - 732}}{{732}} = \frac{{{W_B} \times {M_A}}}{{{M_B} \times {W_A}}} \cr & \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{28}}{{732}} = \frac{{6.5 \times 18}}{{{M_B} \times 100}} \cr & \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{M_B} = 30.6 \cr & \therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta {T_b} = 0.52 \times \frac{{6.5 \times 1000}}{{30.6 \times 100}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1.10 \cr & \therefore \,\,{\text{Boiling point}} = 100 + 1.10 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {101.1^ \circ }C \approx {101^ \circ }C \cr} $$

$$\Delta {T_f} = {K_f} \times m \times i.$$     Since $${K_f}$$  has different values for different solvents, hence even if $$m$$ is the same $$\Delta {T_f}$$  will be different

From the given data
$${P_{{H_2}}} = 1500 \times 0.80$$
$$ = 1200\,mm\,{\text{of}}\,Hg = \frac{{1200}}{{760}}$$      $${\text{atmosphere}} = 1.58\,{\text{atmosphere}}$$
If 's are volumes of gas dissolved by same volume of liquid, then from Henry's law
$$\eqalign{ & \frac{{{V_2}}}{{{V_1}}} = \frac{{{P_2}}}{{{P_1}}};\,\,{\text{or}}\,\,\frac{{{V_2}}}{{20}} = \frac{{1.58}}{{1.0}} \cr & {\text{or}}\,\,{V_2} = 31.60\,mL \cr} $$

Mixture of bromoethane and chloroethane is an ideal solution and will follow Raoult's law as intermolecular interactions of $$A-A$$  and $$B-B$$  type are nearly same as $$A-B$$  type interactions.

Solute in $$200\,g$$  of $$30\% $$  solution $$ = 60\,g$$
Solute in $$300\,g$$  of $$20\% $$  solution $$ = 60\,g$$
Total grams of solute $$ = 120\,g$$
Total grams of solution $$ = 200 + 300 = 500\,g$$
$$\% $$  of solute in the final solution $$ = \frac{{120}}{{500}} \times 100 = 24\% $$

For an ideal solution
(i) There will be no change in volume on mixing the two components i.e. $$\Delta {V_{{\text{mixing}}}} = 0$$
(ii) There will be no change in enthalpy so $$\Delta {H_{{\text{mixing}}}} = 0$$
So, $$\Delta {S_{{\text{mix}}}} \ne 0$$   for an ideal solution.

Real gases show ideal gas behaviour at high temperatures and low pressures.

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

$${\text{Depression in freezing point,}}\,\Delta {T_f} = i \times {K_f} \times m$$
$${\text{Van't}}\,{\text{Hoff factor,}}\,{\text{i = }}\frac{{1 - \alpha + n\alpha }}{1},$$       where $$n =$$  no. of ions produced by complete dissociation of 1 mole of $$HX.$$
$$\eqalign{ & HX \rightleftharpoons {H^ + } + {X^ - } \Rightarrow n = 2 \cr & \therefore \,\,i = \frac{{1 - 0.2 + 2 \times 0.2}}{1} = 1.2\, \cr & \left[ {{\text{For}}\,20\% \,\,{\text{ionisation,}}\,\alpha = \frac{{20}}{{100}} = 0.2} \right] \cr & \therefore \,\Delta {T_f} = 1.2 \times 1.86 \times 0.2 = 0.45\,\,\left[ {\because \,m = 0.2} \right] \cr & {\text{Hence freezing point of solution is }}0 - 0.45 = - 0.45 \cr & \left[ {\because \,{\text{F}}{\text{.P}}\,{\text{of}}\,{\text{water = 0}}{\text{.0C}}} \right] \cr} $$