Raoult’s law is valid for ideal solution only. The two components $$A$$ and $$B$$ follows the condition of Raoult’s law only when the force of attraction between $$A$$ and $$B$$ is equal to the force of attraction between $$A-A$$  and $$B-B.$$
While non-ideal solutions exhibit either positive or negative deviations from Raoult’s law. When $$A - B$$  attraction force is greater than $$A-A$$  and $$B-B,$$  then solution shows negative deviation and when $$A-B$$  attraction force is less than $$A-A$$  and $$B-B,$$  the solution shows positive deviation.

$$\eqalign{ & {\text{Molarity}}\,\,{\text{of}}\,\,N{a_2}C{O_3}\,\,{\text{solution}} \cr & = \frac{{2.65}}{{106}} \times \frac{{1000}}{{250}} \cr & = 0.1\,M \cr & 10\,mL\,\,{\text{of this solution is diluted to}}\,\,500\,mL. \cr & {M_1}{V_1} = {M_2}{V_2} \cr & 0.1 \times 10 = {M_2} \times 500 \Rightarrow {M_2} = 0.002\,M \cr} $$

According to Raoult's Law
$$\frac{{{P^ \circ } - {P_s}}}{{{P_s}}} = \frac{{{W_B} \times {M_A}}}{{{M_B} \times {W_A}}}\,\,\,\,\,\,....({\text{i}})$$
Here $${P^o}{\text{ = }}$$  Vapour pressure of pure solvent,
$${P_s} = $$  Vapour pressure of solution
$${W_B} = $$  Mass of solute, $${W_A} = $$  Mass of solvent
$${M_B} = $$  Molar mass of solute, $${M_A} = $$  Molar Mass of solvent
Vapour pressure of pure water at $${100^ \circ }C$$  ( by assumption = 760 $$torr$$  )
By substituting values in equation (i) we get,
$$\frac{{760 - {P_s}}}{{{P_s}}} = \frac{{18 \times 18}}{{180 \times 178.2}}\,\,\,...{\text{(ii)}}$$
On solving (ii) we get
$${P_s} = 752.4\,torr$$

$$\eqalign{ & \Delta {T_f} = i.{K_f}.m\,;\,0.0054 = i \times 1.8 \times 0.001 \cr & i = 3\,\,{\text{so}}\,{\text{it}}\,{\text{is}}\,\,\left[ {Pt{{\left( {N{H_3}} \right)}_4}C{l_2}} \right]C{l_2}. \cr} $$

$$\eqalign{ & {\text{As T increase, V}}{\text{.P}}{\text{. increases}} \cr & \Delta {T_f} = {K_f} \times m\,\left( {m = {\text{molality of solute}}} \right) \cr & 273 - T{'_f} = 2 \times \frac{{34.5 \times 1000}}{{46 \times 500}} \cr & \therefore \,\,T{'_f} = 270\,K \cr} $$
Thus freezing point of solution $$= 270K.$$  Further as T increases, vapour pressure increases. Hence these facts coincide with the curve given in (C).

Due to association or dissociation of solute molecules there is a change in number of particles. Since colligative properties depend on number of particles there is a change in molecular mass.

$$\eqalign{ & Mole{\text{ fraction of}}\,P = \frac{3}{{3 + 2}} = \frac{3}{5} \cr & Mole{\text{ fraction of}}\,Q = \frac{2}{{3 + 2}} = \frac{2}{5} \cr} $$
Hence, total vapour pressure = $$Mole$$  fraction of $$P$$  ×  vapour pressure of $$P$$  +  $$mole$$  fraction of $$Q$$  ×  vapour pressure of $$Q$$
$$\eqalign{ & = \left( {\frac{3}{5} \times 80 + \frac{2}{5} \times 60} \right) \cr & = 48 + 24 \cr & = 72\,torr \cr} $$

Amalgam of mercury with sodium is an example of solid solution in which mercury (liquid) is solute and sodium (solid) is a solvent.

Powdered sugar has large surface area and thus, dissolves faster in hot water. As solution feels cool to touch, dissolution is endothermic so, dissolution will be favoured at high temperature.

Due to osmosis water moves into the tissues and intercellular spaces causing retention of water.