$$10\,g$$  of glucose is present in $$100\,g$$  of solution Mass of $${H_2}O = 100 - 10 = 90\,g$$
Molality $$ = \frac{{10}}{{180}} \times \frac{{1000}}{{90}}$$
$$ = 0.617\,m$$

Vapour pressure of solution containing $$1\,mole$$  of $$A + 3\,moles$$   of $$B = 550\,\,mm\,\,Hg$$
Vapour pressure of solution containing $$1\,mole$$  of $$A + 4\,moles$$   of $$B = \left( {550 + 10} \right) = 560\,\,mm\,\,Hg$$
$$\eqalign{ & {P_{{\text{Total}}}} = p_A^ \circ \times {x_A} + p_B^ \circ \times {x_B}\,\,{\text{or}}\,\,550 = p_A^ \circ \times {x_A} + p_B^ \circ \times {x_B} \cr & = p_A^ \circ \times \frac{1}{4} + p_B^ \circ \times \frac{3}{4}\,\,\,\,\left[ {\because \,\,{x_A} = \frac{1}{{1 + 3}} = \frac{1}{4},{x_B} = \frac{3}{{1 + 3}} = \frac{3}{4}} \right] \cr & 550 = \frac{{p_A^ \circ }}{4} + \frac{3}{4} \times p_B^ \circ \,\,\,{\text{or}}\,\,\,2200 = p_A^ \circ + 3p_B^ \circ \,\,\,\,\,\,\,\,\,\,\,...\left( {\text{i}} \right) \cr & {\text{Again, we have}} \cr & 560 = p_A^ \circ \times \frac{1}{5} + p_B^ \circ \times \frac{4}{5}\,\,\,\left( {\because \,\,{x_A} = \frac{1}{{1 + 4}} = \frac{1}{5},{x_B} = \frac{4}{{1 + 4}} = \frac{4}{5}} \right) \cr & 2800 = p_A^ \circ + 4p_B^ \circ \,\,\,\,\,\,\,\,...\left( {{\text{ii}}} \right) \cr & {\text{Solving equations (i) and (ii), we get}} \cr & p_B^ \circ = 600\,\,mm\,\,Hg;\,\,p_A^ \circ = 400\,\,mm\,\,Hg \cr} $$

For an ideal solution, $$\,\Delta {H_{{\text{mixing}}}} = 0$$
$$\Delta H = \Delta {H_1} + \Delta {H_2} + \Delta {H_3}$$
( According to Hess's law )
i.e., for ideal solutions there is no change in magnitude of the attractive forces in the two components present.

$$\eqalign{ & \left( i \right)\,\,i = \frac{{{\text{No}}{\text{. of particles after ionisation}}}}{{{\text{No}}{\text{. of particles before ionisation}}}} \cr & \left( {ii} \right)\,\Delta {T_b} = i \times {K_b} \times m \cr & CuC{l_2} \to C{u^{2 + }} + 2C{l^{ - 1}} \cr & \,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \cr & \left( {1 - \alpha } \right)\,\,\,\,\,\,\,\,\,\,\alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2 \cr & i = \frac{{1 + 2\alpha }}{1};\, = 1 + 2\alpha \cr & {\text{Assuming}}\,100\% \,{\text{ionization}} \cr & {\text{So,}}\,\,i = 1 + 2 = 3 \cr & \Delta {T_b} = 3 \times 0.52 \times 0.1 \cr & = 0.156 \approx 0.16 \cr & \left( {m = \frac{{13.44}}{{134.4}} = 0.1} \right) \cr} $$

$$\eqalign{ & {\text{Sodium sulphate dissociates as }} \cr & N{a_2}S{O_4}\left( s \right) \to 2N{a^ + } + SO_4^{ - - } \cr & {\text{hence van't}}\,{\text{hoff}}\,{\text{factor}}\,\,i\, = \,3 \cr} $$
$${\text{Now}}\,\Delta {T_f} = i\,{k_f}.m = 3 \times 1.86 \times 0.01$$       $$ = 0.0558\,K$$

Gaseous densities of ethanol and dimethyl ether would be same at same temperature and pressure. The heat of vaporisation, $$V.P.$$  and $$b.pts$$  will differ due to $$H-$$ bonding in ethanol.

$$\eqalign{ & \left[ {Pt{{\left( {N{H_3}} \right)}_6}} \right]C{l_4} \to {\left[ {Pt{{\left( {N{H_3}} \right)}_6}} \right]^{4 + }} + 4C{l^ - }; \cr & i = 1 + 4 = 5\,{\text{(maximum)}} \cr} $$

$$\eqalign{ & \Delta {T_f} = i \times {K_f} \times m \cr & {\text{Given}}\,\Delta {T_f} = 2.8,{K_f} = 1.8\,K\,kg\,mo{l^{ - 1}}\,\,i = 1 \cr & {\text{( ethylene glygol is a non - electrolyte )}} \cr & {\text{wt}}{\text{. of solvent = 1kg; }}\,\,\,\,{\text{Let of wt of solute = x}} \cr & {\text{Mol}}{\text{. wt of ethylene glycol = 62}} \cr & 2.8 = 1 \times 1.86 \times \frac{x}{{62 \times 1}}\,{\text{or}}\,x = \frac{{2.8 \times 62}}{{1.86}} = 93\,gm \cr} $$

$$\Delta {T_f}$$  ( freezing point depression ) is a colligative property and depends upon the van't Hoff factor $$(i),$$ i.e. number of ions given by the electrolyte in aqueous solution.
$$\Delta {T_f} = i \times {k_f} \times m$$
where, $${k_f} = $$  molal freezing point depression constant
              $$m=$$  molality of the solution
$$\eqalign{ & \therefore {k_f}\,\,{\text{and }}m{\text{ are constant,}} \cr & \therefore \,\,{T_f} \propto i \cr} $$
$${\text{van't Hoff factor for ionic solution}}{\text{.}}$$
$$\left( {\text{A}} \right)KCl\left( {aq} \right) \rightleftharpoons {K^ + }\left( {aq} \right) + C{l^ - }\left( {aq} \right),$$       $$\left( {{\text{Total ions}} = 2\,{\text{thus,}}\,i = 2} \right)$$
$$\left( {\text{B}} \right){C_6}{H_{12}}{O_6} \rightleftharpoons {\text{no ions}}\left[ {i = 0} \right]$$
$${\text{because glucose does not gives ions}}{\text{.}}$$
$$\left( {\text{C}} \right)A{l_2}{\left( {S{O_4}} \right)_3}\left( {aq} \right) \rightleftharpoons 2A{l^{3 + }} + 3SO_4^{2 - }$$        $$\left[ {{\text{Total ions}}\, = 5{\text{,}}\,{\text{thus,}}\,i = 5} \right]$$
$$\left( {\text{D}} \right){K_2}S{O_4}\left( {aq} \right) \rightleftharpoons 2{K^ + } + SO_4^{2 - }$$       $$\left[ {{\text{Total ions}}\, = 3{\text{,}}\,{\text{thus,}}\,i = 3} \right]$$
Hence, $$A{l_2}{\left( {S{O_4}} \right)_3}$$   will exhibit largest freezing point depression due to the highest value of $$i.$$

$$\eqalign{ & {\text{Given}}\,w = 10\,g,\,{\text{Mol}}{\text{. mass}} = 40\,g \cr & {\text{Weight of solvent}} = 1250 \times 0.8\,g = 10000\,g \cr & = 1\,kg \cr & \therefore \,\,{\text{molality}} = \frac{{10}}{{40 \times 1}} = 0.25\,m \cr} $$