$$\eqalign{ & \Delta {T_b} = {K_b}\frac{w}{M} \times \frac{{1000}}{W} \cr & 0.52 = 0.6 \times \frac{3}{m} \times \frac{{1000}}{{200}}\left( {W = 200 \times 1)} \right) \cr & m = \frac{{1.8 \times 5}}{{0.52}} = 17.3\,g\,mo{l^{ - 1}} \cr} $$

Lesser the intermolecular forces, the more the volatile character.

A solution of acetone in ethanol shows positive deviation from Raoult's law. It is because ethanol molecules are strongly hydrogen bonded. When acetone is added, these molecules break the hydrogen bonds and ethanol becomes more volatile. Therefore its vapour pressure is increased.

NOTE : Positive deviations are shown by such solutions in which solvent-solvent and solute-solute interactions are stronger than the solvent interactions. In such solution, the interactions among molecules becomes weaker. Therefore their escaping tendency increases which results in the increase in their partial vapour pressures.
In a solutions of benzene and methanol there exists inter molecular $$H$$ - bonding.

In this solution benzene molecules come between ethanol molecules which weaken intermolecular forces. This results in increase in vapour pressure.

$$\eqalign{ & {\text{Molarity}} = \frac{{1.4275 \times 1000}}{{285.5 \times 250}} = 0.02\,M,\,\pi = CRT \cr & \therefore \,\,{\pi _{cal.}} = 0.02 \times 0.0821 \times 300 = 0.4926\,\,atm \cr & \frac{{{\pi _{obs}}}}{{{\pi _{cal}}}} = i = \frac{{1.478}}{{0.4926}} = 3 \cr & \because \,\,\alpha = \frac{{i - 1}}{{n - 1}} \Rightarrow \alpha = 1\,\,\,\,\left( {\because n = 3} \right) \cr} $$
\[\underset{\begin{smallmatrix} \text{Initial}\,\text{:} \\ \\ \text{Eq}\text{. conc}\text{. :} \end{smallmatrix}}{\mathop{\text{Also},}}\,\,\,\,\,\,\underset{\begin{smallmatrix} 0.02\,M \\ 0 \end{smallmatrix}}{\mathop{\left[ Cr{{\left( N{{H}_{3}} \right)}_{6}} \right]}}\,S{{O}_{4}}.Cl\rightleftharpoons \underset{\begin{smallmatrix} 0 \\ 0.02\,M \end{smallmatrix}}{\mathop{{{\left[ Cr{{\left( N{{H}_{3}} \right)}_{6}} \right]}^{3+}}}}\,+\underset{\begin{smallmatrix} 0 \\ 0.02\,M \end{smallmatrix}}{\mathop{SO_{4}^{2-}}}\,+\underset{\begin{smallmatrix} 0 \\ 0.02\,M \end{smallmatrix}}{\mathop{Cl_{{}}^{-}}}\,\]

$$\eqalign{ & \Delta {T_f} = {K_f} \times m \times i\,; \cr & \Delta {T_f} = 1.85 \times 0.2 \times 1.3 = {0.480^ \circ }C \cr & \therefore \,\,{T_f} = 0 - {0.480^ \circ }C = - {0.480^ \circ }C \cr & \left( {\mathop {HX}\limits_{1 - 0.3} \rightleftharpoons \mathop H\limits_{0.3}^ + + \mathop {{X^ - }}\limits_{0.3} ,i = 1.3} \right) \cr} $$

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

TIPS/Formulae : The salt that ionises to least extent will have highest freezing point. $$\left[ {{\text{i}}{\text{.e}}{\text{., minimum}}\,\Delta {T_f}} \right]$$
Glucose, being non electrolyte, gives minimum no. of particles and hence minimum  $$\Delta {T_f}$$  or maximum $$F.$$ $$pt$$

According to Raoult's law the relative lowering of vapour pressure is equal to mole fraction of solute, i.e. the ratio of number of moles of solute to total number of moles of all component in solution.

The vapour pressure increases with decrease in intermolecular forces. When the forces are weak, the liquid has high volatility and maximum vapour pressure. Diethyl ether has highest vapour pressure while water has lowest vapour pressure.