Solutions MCQ Questions & Answers in Physical Chemistry | Chemistry

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

$$\eqalign{ & {\pi _1} = \frac{{nR{T_1}}}{{{V_1}}},{\pi _2} = \frac{{nR{T_2}}}{{{V_2}}} \cr & {\text{Hence,}}\,\,\frac{{{\pi _1}}}{{{\pi _2}}} = \frac{{580}}{{165}} = \frac{{{V_2}{T_1}}}{{{V_1}{T_2}}} = \frac{{{V_2} \times 290}}{{{V_1} \times 330}} \cr & \Rightarrow \frac{{{V_2}}}{{{V_1}}} = 4 \cr} $$

Mixture of methanol and acetone shows positive deviation from Raoults law. In pure methanol, molecules are hydrogen bonded. On adding acetone, its molecules get in between the host molecules and break some of the hydrogen bonds between them.

$$\eqalign{ & Moles{\text{ of}}\,C{O_2}\,{\text{dissolved}} \cr & = K \times {P_{C{O_2}}} \times {V_{bottle}} = 3 \times {10^{ - 2}} \times 4 \times 100 = 12; \cr & Mass{\text{ of}}\,C{O_2} = 12 \times 44 = 528\,g \cr} $$

$$\Delta {T_f} = i{K_f}\,m$$
For glucose, $$i = 1$$
For $$MgC{l_2},i = 3$$
Thus, the depression in freezing point $$\left( {\Delta {T_f}} \right)$$  of a $$0.01\,M\,MgC{l_2}$$    solution is about three times of that of $$0.01M$$  glucose solution.

For an ideal solution,
$$\left( {\text{i}} \right)\Delta G < 0;$$   for mixing
$$\left( {{\text{ii}}} \right)\Delta {S_{{\text{system}}}} > 0;$$    because disorder increases
$$\left( {{\text{iii}}} \right)\Delta {S_{{\text{surr}}}} = 0;$$    no heat is exchanged in case of ideal solution.
$$\left( {{\text{iv}}} \right)\Delta {H_{{\text{mix}}}} = 0$$

Both $$NaCl$$  and $$KCl$$  have different molecular masses.

$$\eqalign{ & {\text{In case of dissociation}}\,{\text{:}}\,\,\,\,\,\,{A_n}\,\,\,\,\,\, \to \,\,\,\,nA \cr & {\text{Initial number of moles}}\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \cr & {\text{After dissociation}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 - \alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,n\alpha \cr & {\text{No}}{\text{. of particles}} = 1 - \alpha - n\alpha \cr & i = \frac{{1 - \alpha + n\alpha }}{1}\,\,\,{\text{or}}\,\,\,i = 1 - \alpha + n\alpha \cr & i - 1 = n\alpha - \alpha = \left( {n - 1} \right)\alpha \cr & \therefore \,\,\,\alpha = \frac{{i - 1}}{{n - 1}} \cr} $$

$$\eqalign{ & {\text{Vapour pressure of solution}} = {P_A} + {P_B} \cr & = P_A^ \circ {X_A} + P_B^ \circ {X_B} \cr & = \frac{{120 \times 2}}{5} + \frac{{180 \times 3}}{5} \cr & = 156\,mm\,Hg \cr} $$

No. of moles of urea $$ = 0.25\,moles$$
Mass of solvent $$ = 1\,kg = 1000\,g$$
Molar mass of urea $$ = 60\,g\,mo{l^{ - 1}}$$
Mass of urea in $$1000\,g$$  of $${H_2}O = 0.25 \times 60 = 15\,g$$
Total mass of solution $$ = 1000 + 15 = 1015\,g$$
$$1.015\,kg$$   solution contains $$15\,g$$  of urea
$$2.5\,kg$$   solution will contain $$ = \frac{{15}}{{1.015}} \times 2.5 = 37\,g$$     of urea