Solutions MCQ Questions & Answers in Physical Chemistry | Chemistry

$$\eqalign{ & \Delta {T_f} = i{K_f}m \cr & KBr \to \,\,{K^ + } + B{r^ - } \cr & 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \cr & \left( {1 - \alpha } \right)\,\,\,\,\,\alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\alpha \cr & i = 1 + \alpha \cr & {\text{Here}}\,\,\alpha = 0.8\,\,\left( {80\% \,{\text{ionization}}} \right) \cr & = 1.8 \cr & \Delta {T_f} = \left( {1.8} \right)\left( {1.86} \right)\left( {0.4} \right) = 1.339 \cr & {T_f} = - {1.339^ \circ }C \cr} $$

When acetone and chloroform are mixed together, hydrogen bonds are formed between them which increases intermolecular interactions hence decreases the vapour pressure showing negative deviation.

The given solution is unsaturated ( saturated one contains $$5\,g$$  of solute in $$50\,g$$  of water ). On evaporation, concentration of solution increases to saturation and thereafter becomes constant due to excess solid separating out. Hence, osmotic pressure first increases and then becomes constant.

$$\eqalign{ & {V_1} = 30\,mL,{M_1} = 0.5\,M,{V_2} = 500\,mL,{M_2} = ? \cr & {M_1}{V_1} = {M_2}{V_2} \cr & 0.5 \times 30 = {M_2} \times 500\,\,\,{\text{or}}\,\,\,{{\text{M}}_2} = 0.03\,M \cr} $$

$$\eqalign{ & \Delta {T_f} = {K_f} \times m \cr & M = \frac{{1000 \times {K_f} \times {w_2}\left( {{\text{solute}}} \right)}}{{\Delta {T_f} \times {w_1}\left( {{\text{solvent}}} \right)}} \cr & = \frac{{1000 \times 1.86 \times 1.8}}{{0.465 \times 40}} \Rightarrow M = 180 \cr & {\text{Molecular formula = (empirical formula}}{{\text{)}}_n} \cr & n = \frac{{{\text{Molecular mass}}}}{{{\text{Empirical formula mass}}}} = \frac{{180}}{{30}} = 6 \cr & {\text{Molecular formula}} = {\left( {C{H_2}O} \right)_6} = {C_6}{H_{12}}{O_6}. \cr} $$

$$\eqalign{ & \Delta {T_b} = i{K_b}\,m \cr & {T_b} - T_b^ \circ = i{K_b}\,m \cr} $$
Thus, boiling point of solution $$\left( {{T_b}} \right)$$  depends on value of van't Hoff factor $$(i).$$
For $$1.0\,M\,N{a_2}S{O_4}$$   solution, $$i=3$$  hence, it has highest boiling point.

$$\eqalign{ & {P_{{N_2}}} = {\kappa _H}{\chi _{{N_2}}};\,0.8 \times 5 = 1 \times {10^5} \times {\chi _{{N_2}}} \cr & \therefore \,\,{\chi _{{N_2}}} = 4 \times {10^{ - 5}};\,{\text{Solubility in 10 moles = }}\,4 \times {10^{ - 4}} \cr} $$

$$\eqalign{ & {\text{Molality}}\left( m \right) = \frac{{1000 \times n}}{{N \times M}} \cr & {\text{where,}} \cr & n = {\text{number of moles of solute}} \cr & N = {\text{number of moles of solvent}} \cr & M = {\text{molar mass of solvent}} \cr & {\text{Given}},\,m = 1 \cr & \therefore \,\,1 = \frac{{1000 \times n}}{{N \times 18}} \cr & \Rightarrow \,\frac{n}{N} = \frac{{18}}{{1000}} \cr & {\text{or}}\,\,\frac{n}{{n + N}} = \frac{{18}}{{1018}} = 0.0177 \cr & {\text{Alternate Method}} \cr & N = \frac{{1000}}{{18}} = 55.5\,mol \cr & \,\,n = 1 \cr} $$
[ $$\therefore $$ $$1$$ $$m$$ solution implies that $$1$$ $$mole$$  of solute is present in $$1$$ $$kg$$  or $$1000$$  $$g$$  water ]
$$\eqalign{ & \therefore {\text{Mole fraction of solute}} \cr & = \frac{n}{{n + N}} \cr & = \frac{1}{{1 + 55.5}} \cr & = 0.0177 \cr} $$

A saturated solution cannot dissolve any more solute at that temperature. If precipitation occurs, it is supersaturated.

TIPS/FORMULAE :
$$\eqalign{ & {\text{Osmotic pressure}}\,\left( \pi \right)\,{\text{of isotonic solutions are equal}}{\text{.}} \cr & {\text{For solution of unknown substance}}\,\left( {\pi = {\text{CRT}}} \right) \cr & {C_1} = \frac{{\frac{{5.25}}{M}}}{V} \cr & {\text{For solution of urea}},\,{C_2}\left( {{\text{concentration}}} \right) = \frac{{\frac{{1.5}}{{60}}}}{V} \cr & {\text{Given,}}\,\,{\pi _1} = {\pi _2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\because \pi = CRT \cr & \therefore \,\,{C_1}RT = {C_2}RT\,\,{\text{or}}\,\,{C_1} = {C_2}\,\,{\text{or}}\,\,\frac{{\frac{{5.25}}{M}}}{V} = \frac{{\frac{{1.8}}{{60}}}}{V} \cr & \therefore \,M = 210g/mol \cr} $$