$$\eqalign{ & {\text{As}}\,\,\Delta {T_f} = {K_f}m \cr & \Delta {T_b} = {K_b}.\,m \cr & {\text{Hence, we have}}\,m{\text{ = }}\frac{{\Delta {T_f}}}{{{K_f}}} = \frac{{\Delta {T_b}}}{{{K_b}}} \cr & {\text{or}}\,\,\Delta {T_f} = \Delta {T_b}\frac{{{K_f}}}{{{K_b}}} \cr & \Rightarrow \left[ {\Delta {T_b} = 100.18 - 100 = {{0.18}^ \circ }C} \right] \cr & = 0.18 \times \frac{{1.86}}{{0.512}} \cr & = {0.654^ \circ }C \cr} $$
As the freezing point of pure water is $${0^ \circ }C,$$
$$\eqalign{ & \Delta {T_f} = 0 - {T_f} \cr & 0.654 = 0 - {T_f} \cr & \therefore \,\,{T_f} = - 0.654 \cr} $$
Thus the freezing point of solution will be $$ - {\text{0}}{\text{.65}}{{\text{4}}^ \circ }C.\,$$

Key Idea For this problem, the following expression can be used.
$$\eqalign{ & \Delta {G_{{\text{mix}}}} = \Delta {H_{{\text{mix}}}} - T\Delta {S_{{\text{mix}}}} \cr & {\text{For an ideal gas}} \cr & \Delta {H_{{\text{mix}}}} = 0;\,\,\Delta {U_{{\text{mix}}}} = 0;\,\,\Delta {S_{{\text{mix}}}} \ne 0 \cr} $$
Putting all these values in the expression,
$$\Delta {G_{{\text{mix}}}} = \Delta {H_{{\text{mix}}}} - T\Delta {S_{{\text{mix}}}}$$       $$ \Rightarrow \Delta {G_{{\text{mix}}}} = 0 - T\Delta {S_{{\text{mix}}}}$$
$$\therefore \,\,\Delta {G_{{\text{mix}}}} \ne 0$$
Thus, option (D) is incorrect.

Solubility changes with temperature. A plant cell shrinks in hypertonic solution. Relative lowering of vapour pressure is a colligative property.

According to Henry’s law the mass of a gas dissolved per unit volume of solvent is proportional to the pressure of the gas at constant temperature $$m = Kp$$   i.e. as the solubility increases, value of Henry’s law constant decreases. Since $$C{O_2}$$  is most soluble in water among the given set of gases. Therefore $$C{O_2}$$  has the lowest value of Henry’s law constant.

$$\eqalign{ & {\text{For non electrolyte solute (glucose)}}\,i = 1 \cr & {\text{For}}\,\,BaC{l_2} \rightleftharpoons B{a^{2 + }} + 2C{l^ - } \cr & i = 3\,\left( {100\% \,{\text{ionized}}} \right) \cr} $$
$$i = \left( {{\text{number of ions after}}\,\,{\text{ionization}} \times \alpha } \right)$$        $$ + \left( {1 - \alpha } \right)$$
$$\eqalign{ & = \left( {3 \times 1} \right) + \left( {1 - 1} \right) \cr & = 3 \cr & \pi = \left( {i{C_1} - {i_2}{C_2}} \right)RT \cr & \Rightarrow \left( {1 \times 0.2 - 3 \times 0.05} \right)0.0821 \times 300 \cr & \Rightarrow 1.23\,atm \cr} $$

The solubility of the gas in liquids decreases with the increase in value of $$KH$$ at a given pressure.

$${P_{{\text{total}}}} = P_A^o{X_A} + P_B^o{X_B}\,;$$     $$550 = P_A^o \times \frac{1}{4} + P_B^o \times \frac{3}{4}$$
$$\eqalign{ & P_A^o + 3P_B^o = 550 \times 4\,...\left( {\text{i}} \right) \cr & {\text{In second case}} \cr & {P_{{\text{total}}}} = P_A^o \times \frac{1}{5} + P_B^o \times \frac{4}{5} \cr & P_A^o + 4P_B^o = 560'5\,\,\,\,\,\,...\left( {{\text{ii}}} \right) \cr & {\text{Subtract (i) from (ii)}} \cr} $$
$$\therefore \,\,P_B^o = 560 \times 5 - 550 \times 4 = 600$$         $$\because P_A^o = 400$$

$$\eqalign{ & \Delta {T_f} = i \times {K_f} \times m \cr & 0.0054 = i \times 1.8 \times 0.001 \Rightarrow i = 3 \cr} $$
Since it gives 3 particles on dissociation the correct formula of the molecule is $$\left[ {Pt{{\left( {N{H_3}} \right)}_4}C{l_2}} \right]C{l_2}.$$

TIPS/Formulae :
$$\eqalign{ & (i)\,i = \frac{{{\text{No}}{\text{. of particles after ionisation }}}}{{{\text{No}}{\text{. of particles before ionisation}}}} \cr & (ii)\,\Delta {T_b} = i \times {K_b} \times m \cr} $$

\[\underset{\begin{smallmatrix} 1 \\ \left( 1-\alpha \right) \end{smallmatrix}}{\mathop{CuC{{l}_{2}}}}\,\to \underset{\begin{smallmatrix} 0 \\ \alpha \end{smallmatrix}}{\mathop{C{{u}^{2+}}}}\,+\underset{\begin{smallmatrix} 0 \\ 2\alpha \end{smallmatrix}}{\mathop{2C{{l}^{-}}}}\,\]
$$\eqalign{ & i = \frac{{1 + 2\alpha }}{1}\,\,i = 1 + 2\alpha \cr & {\text{Assuming }}100\% {\text{ ionization}}\,\,\,\,\,\,\,\,\,So,\,i = 1 + 2 = 3 \cr & \Delta {T_b} = 3 \times 0.52 \times 0.1 = 0.156 \approx 0.16\,\,\,\left[ {m = \frac{{13.44}}{{134.4}} = 0.1} \right] \cr} $$

If a pressure higher than the osmotic pressure is applied on the solution, then reverse osmosis takes place i.e., the solvent will flow from the solution ( i.e., high conc. of solute ) into the pure solvent ( i.e., low conc. of solute ) through the semipermeable membrane.