Solutions MCQ Questions & Answers in Physical Chemistry | Chemistry

Both $${K_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right]$$   and $$A{l_2}{\left( {S{O_4}} \right)_3}$$   gives 5 ions after dissociation, so they have same value of van't Hoff factor $$(i).$$
$$\eqalign{ & {K_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right] \to 4{K^ + } + {\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{4 - }} \cr & A{l_2}{\left( {S{O_4}} \right)_3} \to 2A{l^{3 + }} + 3SO_4^{2 - } \cr} $$

From the relation, $$\Delta {{\text{T}}_f} = i{K_f}m,$$    it is evident that $$\Delta {T_f} \propto i.$$   Since $$i$$ is maximum for ethanol ( others are non electrolytes ), $$\Delta {T_f}$$  ( depression in freezing point ) will be maximum for $${C_2}{H_5}OH$$   and hence freezing point will be minimum for ethanol solution.

Due to stronger intermolecular interactions in acetone and chloroform lesser number of molecules vaporise resulting in low vapour pressure and high boiling point.

$$p = {K_H}x$$
Higher the value of $${K_H}$$  at a given pressure, the lower is the solubility of the gas in the liquid.

\[\begin{align} & \underset{\begin{smallmatrix} \text{Strength}\left( g\,{{L}^{-1}} \right): \\ \text{Molarity:} \\ i: \end{smallmatrix}}{\mathop{{}}}\,\,\,\,\,\,\,\,\underset{\begin{smallmatrix} 34.2 \\ \frac{34.2}{342}=0.1 \\ 1 \end{smallmatrix}}{\mathop{\text{Sucrose}}}\,\,\,\,\,\,\,\,\,\,\,\,\text{ }\underset{\begin{smallmatrix} 60 \\ \frac{60}{60}=1 \\ 1 \end{smallmatrix}}{\mathop{\text{Urea}}}\,\text{ }\,\,\,\,\,\,\,\,\,\,\,\underset{\begin{smallmatrix} 90 \\ \frac{90}{180}=0.5 \\ 1 \end{smallmatrix}}{\mathop{\text{Glucose}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\underset{\begin{smallmatrix} 58.5 \\ \frac{58.5}{58.5}=1 \\ 2 \end{smallmatrix}}{\mathop{NaCl}}\, \\ & \\ \end{align}\]
We know, $$\pi = iCRT$$
Thus, osmotic pressure of these solutions will lie in following sequence :
$$\mathop {{\text{Sucrose }}}\limits_{\left( {\text{i}} \right)} {\text{ < }}\mathop {{\text{Glucose}}}\limits_{\left( {{\text{iii}}} \right)} {\text{ < }}\mathop {{\text{Urea }}}\limits_{\left( {{\text{ii}}} \right)} {\text{ < }}\mathop {NaCl}\limits_{\left( {{\text{iv}}} \right)} $$

Benzoic acid exists as dimer in benzene.

[ Normal molecular mass = 122 amu observed molecular mass = 244 amu, in case of complete association ]
$$\alpha = \frac{{20}}{{100}} = 0.2\,\,\,\alpha = \frac{{20}}{{100}} = 0.2\,\,\,\alpha = \frac{{20}}{{100}} = 0.2$$

Abnormality present due to dissociation and association of the solution.
For dissociation, van’t Hoff factor is greater than one and for association, van’t Hoff factor is less than one.
For dissociation, $$i>1$$
For association, $$i<1$$

According to depression in freezing point,
$$\eqalign{ & \Delta {T_f} = i \times {k_f}.\,m \cr & {\text{where,}}\,\,{k_f} = {\text{cryoscopic constant}} \cr & {\text{or}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,i = \frac{{\Delta {T_f} \times {W_{{\text{solvent}}}}}}{{{k_f} \times {n_{{\text{solute}}}} \times 1000}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{3.82 \times 45}}{{1.86 \times \left( {\frac{5}{{142}}} \right) \times 1000}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2.63 \cr} $$

$$\eqalign{ & \Delta {T_f} = i.{K_f}.m\,;\,\Delta {T_b} = i.{K_b}.m \cr & \frac{{\Delta {T_f}}}{{\Delta {T_b}}} = \frac{{{K_f}}}{{{K_b}}} \cr & \Delta {T_f} = 0 - \left( { - {{0.186}^ \circ }C} \right) = {0.186^ \circ }C \cr & \frac{{0.186}}{{\Delta {T_b}}} = \frac{{1.86}}{{0.52}} \Rightarrow \Delta {T_b} = \frac{{0.52 \times 0.186}}{{1.86}} = 0.052 \cr} $$

\[\begin{align} & \begin{matrix} {} & N{{a}_{2}}S{{O}_{4}} & \to \\ \text{Mol}\text{. before dissociation } & 1 & {} \\ \text{Mol}\text{. after dissociation} & 1-\alpha & {} \\ \end{matrix}\,\,\,\begin{matrix} 2N{{a}^{+}} & + & SO_{4}^{2-} \\ 0 & {} & 0 \\ 2\alpha & {} & 1\alpha \\ \end{matrix} \\ & i=1-\alpha +2\alpha +\alpha =1+2\alpha \\ \end{align}\]