Solutions MCQ Questions & Answers in Physical Chemistry | Chemistry

(iii) and (iv) will form ideal solutions hence do not form azeotropes. Azeotropes have same composition in liquid and vapour form when distilled.

$$\eqalign{ & {\text{Equivalent weight of dibasic acid}} \cr & \,\,\,\,\,\, = \frac{{{\text{Molecular weight}}}}{2} \cr & E = \frac{{200}}{2} \cr & \,\,\,\,\,\, = 100 \cr & {\text{Strength}} = 0.1\,N,\,m = ?\,V = 100\,mL \cr & {\text{Normality}} \cr & \left( N \right) = \frac{{{\text{Mass}}}}{E} \times \frac{{1000}}{{V\left( L \right)}} \cr & \,\,\,\,w = \frac{{E\,NV}}{{1000}} \cr & \,\,\,\,\,\,\,\,\,\, = \frac{{100 \times 100 \times 0.1}}{{1000}} \cr & \,\,\,\,\,\,\,\,\,\, = 1\,g \cr} $$

Lower the $$B.$$ $$pt.,$$ higher will be the $$V.P.$$  The $$V.P.$$  of the mixture is greater than either of the two liquids.
[NOTE : In case of positive deviation from Roult's law the partial vapour pressure of each liquid and total vapour pressure of solution will be greater as compared to initial solution]

$$\eqalign{ & \Delta {T_b} = {K_b}\frac{{{W_B}}}{{{M_B} \times {W_A}}} \times 1000; \cr & \Delta {T_f} = {K_f}\frac{{{W_B}}}{{{M_B} \times {W_A}}} \times 1000; \cr & \frac{{\Delta {T_b}}}{{\Delta {T_f}}} = \frac{{{K_b}}}{{{K_f}}} = \frac{{\Delta {T_b}}}{{ - 0.186}} = \frac{{0.512}}{{1.86}} = {0.0512^ \circ }C. \cr} $$

Total vapour pressure = vapour pressure of pure benzene + vapour pressure of toluene
$$\eqalign{ & = 100 + 50 \cr & = 150\,mm \cr & {\text{We}}\,\,{\text{know,}} \cr & P_{{C_6}{H_6}}^ \circ = P \times {X_{{C_6}{H_6}}} \cr & 100 = 150 \times {X_{{C_6}{H_6}}} \cr & {X_{{C_6}{H_6}}} = \frac{{100}}{{150}} = 0.67 \cr} $$

van’t Hoff factor $$(i)$$  and the degree of association are related as below :
$$\eqalign{ & i = 1 - \alpha \left( {1 - \frac{1}{n}} \right) \cr & 0.9 = 1 - 0.2\left( {1 - \frac{1}{n}} \right) \cr & {\text{On}}\,\,{\text{solving,}} \cr & \left( {1 - \frac{1}{n}} \right) = \frac{1}{2} \cr & \frac{1}{n} = 1 - \frac{1}{2} = \frac{1}{2} \cr & \therefore \,\,n = 2 \cr} $$

$$\eqalign{ & {\text{For dilute solution,}} \cr & pV = nRT\,\,\left( {p = \pi ,\,{\text{osmotic pressure}}} \right) \cr & {\text{or}}\,\,\,\pi V = nRT\,\,\,{\text{or}}\,\,\pi = \frac{n}{V}RT \cr & \Rightarrow \pi V = \frac{{{m_2}}}{{{M_2}}}RT \Rightarrow {M_2} = \frac{{{m_2}Rt}}{{\pi V}} \cr & {\text{where}},\,\pi = {\text{osmotic pressure}} \cr & {\text{V}} = {\text{volume of solution}} \cr & n = {\text{number of moles of solute}} \cr & {m_2} = {\text{mass of solute}} \cr & {M_2} = {\text{molecular mass of solute}} \cr} $$

Added $$Hg{I_2}$$  forms a complex with $$KI$$  in the solution as follows
$$2KI + Hg{I_2} \to {K_2}\left[ {Hg{I_4}} \right]$$
As a result, number of particles decreases and so $$\Delta {T_f}$$  increases.
[ NOTE : : Depression in freezing point is a colligative property ]

NOTE: The salt producing highest number of ions will have lowest freezing point.
$${K_2}S{O_4} \to 2{K^ + } + SO_4^{2 - };\,{K_2}S{O_4}$$       gives highest number of particles ( 2 + 1 = 3 ).
Glucose, being non-electrolyte gives minimum no. of particles and hence minimum $$\Delta {T_f}$$  or maximum $$F.$$ $$pt.$$

According to Nernst distribution law when we mixed a common solute in a pair of immiscible liquids, then the ratio of amount of solute in both liquids is fixed at a fixed temperature.