Solutions MCQ Questions & Answers in Physical Chemistry | Chemistry

$$\eqalign{ & \pi = i\,CRT \cr & ^\pi {C_2}{H_5}OH = 1 \times 0.500 \times R \times T = 0.5RT \cr & ^\pi M{g_3}{\left( {P{O_4}} \right)_2} = 5 \times 0.100 \times R \times T = 0.5\,RT \cr & ^\pi KBr = 2 \times 0.250 \times R \times T = 0.5\,RT \cr & ^\pi N{a_3}P{O_4} = 4 \times 0.125 \times RT = 0.5\,RT \cr} $$
Since the osmotic pressure of all the given solutions is equal. Hence all are isotonic solution.

Hypertonic solution has high osmotic pressure. When a plant cell is placed in hypertonic solution water will diffuse out of the cell resulting in shrinking of the cell.

$$\eqalign{ & {\text{No}}{\text{. of moles of }}NaOH \cr & = \frac{{M \times V}}{{1000}} \cr & = \frac{{0.5 \times 250}}{{1000}} \cr & = 0.125 \cr & {\text{Mass of }}NaOH = 40 \times 0.125 = 5\,g \cr} $$

$$\eqalign{ & 5X \to {X_5} \cr & i = 1 - \left( {1 - \frac{1}{n}} \right) \cr & \,\,\, = 1 - \left( {1 - \frac{1}{5}} \right) = 0.2 \cr} $$

From Raoult law relation,
$$\frac{{{p^ \circ } - p}}{{{p^ \circ }}} = \frac{{{\text{No}}{\text{.of moles of solute}}}}{{{\text{No}}{\text{.of moles of solvent + No}}{\text{. of moles of solute}}}}$$
When the concentration of solute is much lower than the concentration of solvent,
$$\eqalign{ & \frac{{{p^ \circ } - p}}{{{p^ \circ }}} = \frac{{{\text{No}}{\text{.of moles of solute}}}}{{{\text{No}}{\text{.of moles of solvent}}}} \cr & \frac{{760 - p}}{{760}} - \frac{{\frac{{2.5}}{m}}}{{\frac{{100}}{{18}}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....({\text{i}}) \cr} $$
From elevation in boiling point, $$\Delta {T_b} = {K_b} = {K_b} \times m$$
$$\eqalign{ & 2 = 0.76 \times m \cr & m = \frac{2}{{0.76}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....({\text{ii}}) \cr & {\text{From (i) and (ii), }}p = 724\,mm \cr} $$

$${\text{Moles of glucose}}\,\,\frac{{10}}{{180}} = 0.05,$$      $${\text{urea}} = \frac{{10}}{{60}} = 0.16,$$     $${\text{Sucrose}} = \frac{{10}}{{342}} = 0.029$$
$${\text{Hence osmotic presure}}$$      $${P_2} > {P_1} > {P_3}\left( {\pi \propto C} \right)$$

$$\eqalign{ & Mg{\left( {N{O_3}} \right)_2} \to M{g^{2 + }} + 2NO_3^ - \cr & \alpha = \frac{{i - 1}}{{n - 1}} \cr & \,\,\,\, = \frac{{2.74 - 1}}{{3 - 1}} \cr & \,\,\, = \frac{{1.74}}{2} \cr & \,\,\, = 0.87 \cr & {\text{Degree of dissociation}} \cr & = 0.87 \times 100 \cr & = 87\% \cr} $$

No. of moles of $${H_2}S = 0.195$$
No. of moles of $${H_2}O = \frac{{1000}}{{18}} = 55.55\,mol$$
MoIe fraction of $${H_2}S = \frac{{0.195}}{{0.195 + 55.55}} = 0.0035$$
Pressure at $$STP = 0.987\,bar$$
According to Henry's law, $$p = {K_H}x$$
or $${K_H} = \frac{{{p_{{H_2}S}}}}{{{x_{{H_2}S}}}} = \frac{{0.987}}{{0.0035}} = 282\,bar$$

Vapour pressure of pure water at boiling point $$ = 1\,atm = 1.013\,bar$$
Vapour pressure of solution $$\left( {{p_s}} \right) = 1.004\,bar$$
Let mass of solution $$ = 100\,g$$
Mass of solute $$ = 2\,g$$
Mass of solvent $$ = 100 - 2 = 98\,g$$
$$\eqalign{ & \frac{{{p^ \circ } - {p_s}}}{{{p^ \circ }}} = \frac{{{n_2}}}{{{n_1} + {n_2}}} = \frac{{{n_2}}}{{{n_1}}} = \frac{{{W_2}/{M_2}}}{{{W_1}/{M_1}}}\left( {\because \,{n_2} < < < {n_1}} \right) \cr & \frac{{1.013 - 1.004}}{{1.013}} = \frac{2}{{{M_2}}} \times \frac{{18}}{{98}} \cr & {\text{or}}\,\,\,{M_2} = \frac{{2 \times 18}}{{90}} \times \frac{{1.013}}{{0.009}} = 41.35\,g\,mo{l^{ - 1}}. \cr} $$

$$\eqalign{ & {\text{We}}\,{\text{know}},\,\pi - iCRT;\,{\pi _{xy}} = 4{\pi _{BaC{l_2}}} \cr & \therefore \,\,2\left[ {XY} \right] = 4 \times \left( {0.01} \right) \times 3 \cr & \left[ {XY} \right] = 0.06 \cr & = 6 \times {10^{ - 2}}\,mol/L \cr} $$