$$\eqalign{ & {\text{Number of moles in 92 g of N}}{{\text{a}}^ + } = \frac{{92}}{{23}} = 4\,moles \cr & {\text{Molality (m) = }}\frac{{{\text{Number}}\,{\text{of moles}}}}{{{\text{Mass of solvent (in kg)}}}} \cr & \therefore \,\,m = \frac{4}{1} = 4\,mol\,k{g^{ - 1}} \cr} $$

$$\eqalign{ & {\text{Mole fraction of}}\,\,P = \frac{3}{{3 + 2}} = 0.6 \cr & {\text{Mole fraction of}}\,\,Q = \frac{2}{{3 + 2}} = 0.4 \cr & {P_{{\text{total}}}} = {p_P} + {p_Q} = p_P^ \circ {x_P} + p_Q^ \circ {x_Q} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 80 \times 0.6 + 60 \times 0.4 = 72\,torr \cr} $$

$$\frac{{{p^ \circ } - {p_s}}}{{{p^ \circ }}} = \frac{n}{{n + N}}i;$$
$$\Delta P$$  is minimum since $$i= 1$$  for urea. Thus, urea solution will have its vapour pressure closer to solvent.

$$\eqalign{ & {n_{CHC{l_3}}} = \frac{{25.5}}{{119.5}} = 0.213 \cr & {n_{C{H_2}C{l_2}}} = \frac{{40}}{{85}} = 0.47 \cr & {P_T} = P_A^ \circ {X_A} + P_B^ \circ {X_B} \cr & = 200 \times \frac{{0.213}}{{0.683}} + 41.5 \times \frac{{0.47}}{{0.683}} \cr & = 62.37 + 28.55 \cr & = 90.92 \cr} $$

$$\eqalign{ & {\text{Depression in freezing point,}} \cr & \Delta {T_f} = {k_f} \times m \cr & {\text{where,}}\,m = {\text{molality}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{W_B} \times 1000}}{{{M_B}.{W_A}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{68.5 \times 1000}}{{342 \times 1000}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{68.5}}{{342}} \cr & \Delta {T_f} = 1.86 \times \frac{{68.5}}{{342}} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = {0.372^ \circ }C \cr & {T_f} = T_F^0 - \Delta {T_f} \cr & \,\,\,\,\,\,\, = 0 - {0.372^ \circ }C \cr & \,\,\,\,\,\,\, = - {0.372^ \circ }C \cr} $$

$$\eqalign{ & {K_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right] \to 4{K^ + } + {\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{4 - }} \cr & A{l_2}{\left( {S{O_4}} \right)_3} \to 2A{l^{3 + }} + 3SO_4^{2 - } \cr} $$

The solubility of gas in a liquid increases with increase in pressure and is directly proportional to the pressure of the gas.

In solution containing $$A$$ and $$B$$ component showing negative deviation $$A - A$$  and $$B - B$$  interactions are weaker than that of $$A - B$$  interactions. For such solutions.
$$\Delta H = - ve\,\,{\text{and}}\,\,\Delta V = - ve$$

Vapour pressure of a solution containing nonvolatile solute is less than that of the pure solvent. The decrease in vapour pressure depends upon the quantity of non-volatile solute present in it. Hence, vapour pressure of $$A > C > B.$$

We know that, according to Raoult’s law
$$\eqalign{ & \,\,\,\frac{{{p^ \circ } - p}}{{{p^ \circ }}} = {\chi _B} \cr & \frac{{{p^ \circ } - 60}}{{{p^ \circ }}} = \frac{{{n_B}}}{{{n_A} + {n_B}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{0.2}}{{0.2 + 0.8}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{0.2}}{{1.0}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{2}{{10}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{5} \cr & {p^ \circ } - 60 = \frac{{{p^ \circ }}}{5} \cr & \Rightarrow \,\,\,{p^ \circ } - \frac{{{p^ \circ }}}{5} = 60 \cr & \,\,\,\,\,\frac{{5{p^ \circ } - {p^ \circ }}}{5} = 60 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4{p^ \circ } = 60 \times 5 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{p^ \circ } = \frac{{60 \times 5}}{4} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{300}}{4} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 75\,mm\,{\text{of}}\,Hg \cr} $$