Solutions MCQ Questions & Answers in Physical Chemistry | Chemistry

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

Key Idea Strong electrolytes dissociate completely in their solutions.
van't Hoff factor = total number of ions after dissociation
$$\eqalign{ & {\text{So for}}\,\,\,Ba{\left( {OH} \right)_2} \rightleftharpoons B{a^{2 + }}\left( {aq} \right) + 2O{H^ - }\left( {aq} \right) \cr & {\text{Number of ions}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2 \cr & {\text{van't Hoff factor,}}\,\,i = 1 + 2 = 3 \cr} $$

Solutions in which solute - solute and solvent-solvent interactions are almost similar to solute solvent interactions are known as ideal solution.

The bacteria on salted meat or candied fruit loss water through osmosis, shrink and die.

Molarity and normality are temperature dependent because they involve volume of solutions. Volume is dependent on temperature.
$${\text{Molarity}}\left( M \right) = \frac{{{\text{Moles of solute}}}}{{{\text{Volume of solution}}\left( {{\text{in}}\,L} \right)}}$$
Molality, mole fraction and weight percentage does not depend on temperature because they involve masses of solute and solvent.

Given, Vapour pressure of benzene $${\text{ = 75 }}torr$$
Vapour pressure of benezene $${\text{ = 22 }}torr$$
mass of benzene in $${\text{ = }}78g$$
hence moles of benzene $$ = \frac{{78}}{{78}} = 1\,mole$$
( $$mol.$$  $$wt$$  of benzene = 78 )
mass of toluence in solution $${\text{ = 46}}g$$
hence moles of toluene $${\text{ = }}\frac{{46}}{{92}} = 0.5\,mole$$
now partial pressure of benezene
$$ = {P^ \circ }_b.{X_b} = 75 \times \frac{1}{{1 + 0.5}} = 50\,torr$$        $$ = 75 \times \frac{1}{{1.5}} = 75 \times \frac{2}{3} = 50$$

$$\eqalign{ & {x_{{C_2}{H_5}OH}} = \frac{{{n_{{C_2}{H_5}OH}}}}{{{n_{{C_2}{H_5}OH}} + {n_{{H_2}O}}}} \cr & {\text{Mass of }}{C_2}{H_5}OH = 85\,g \cr & {\text{Molar mass of}}\,\,{C_2}{H_5}OH = 46\,g/mol \cr & {n_{{C_2}{H_5}OH}} = \frac{{85}}{{46}} = 1.85\,mol \cr & {\text{Mass of water}} = 100 - 85 = 15\,g \cr & {n_{{H_2}O}} = \frac{{15}}{{18}} = 0.833\,mol \cr & {x_{{C_2}{H_5}OH}} = \frac{{1.85}}{{1.85 + 0.833}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{1.85}}{{2.683}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.69 \cr} $$

$$\eqalign{ & {\text{In benzene}} \cr & \mathop {2C{H_3}COOH}\limits_{1 - \alpha } \rightleftharpoons \mathop {{{\left( {C{H_3}COOH} \right)}_2}}\limits_{\frac{\alpha }{2}} \cr & i = 1 - \alpha + \frac{\alpha }{2} = 1 - \frac{\alpha }{2} \cr & {\text{Here }}\alpha {\text{ is degree of association}} \cr & \Delta {{\text{T}}_f} = i{K_f}m \cr & 0.45 = \left( {1 - \frac{\alpha }{2}} \right)\left( {5.12} \right)\frac{{\left( {\frac{{0.2}}{{60}}} \right)}}{{\frac{{20}}{{1000}}}} \cr & 1 - \frac{\alpha }{2} = 0.527 \cr & \alpha = 0.945 \cr & \% {\text{ degree of association = 94}}{\text{.6}}\% \cr} $$

Acetone + ethyl alcohol solution shows positive deviation while acetone + chloroform shows negative deviation.
Other examples :
Positive deviations - Acetone + ethyl alcohol, acetone + benzene, water + ethyl alcohol
Negative deviations - Nitric acid + water, benzene + chloroform

Colligative properties depends upon the no. of particles. Since methanol is non electrolyte hence cannot be considered.