Solutions MCQ Questions & Answers in Physical Chemistry | Chemistry

$$\eqalign{ & {\text{Mass}}\,\,{\text{of}}\,\,{\text{solution}} \cr & = {\text{Mass}}\,\,{\text{of}}\,\,{C_6}{H_6} + {\text{Mass}}\,\,{\text{of}}\,\,CC{l_4} \cr & = 22 + 122 \cr & = 144\,g \cr & {\text{Mass}}\,\,\% \,\,{\text{of}}\,\,CC{l_4} \cr & = \frac{{122}}{{144}} \times 100 \cr & = 84.72\% \cr} $$

$$\eqalign{ & {\text{Molecular weight of naphthoic acid}} \cr & {C_{11}}{H_8}{O_2} = 172\,gmo{l^{ - 1}}. \cr & {\text{The theoretical value of depression in freezing point}} \cr & {K_f} \times molality = 1.72 \times \frac{{20 \times 1000}}{{172 \times 50}} = 4K \cr & {\text{Van't Hoff factor,}} \cr & i = \frac{{{\text{Observed value of colligative property}}}}{{{\text{Theoretical value of colligative property}}}} \cr & = \frac{2}{4} = 0.5 \cr} $$

According to Raoult's law, if volatile liquid added in pure solvent, then total pressure equal to sum of the partial pressure of volatile liquid and solvent.
$$\eqalign{ & {\text{Total pressure,}}\,\,\,\,{p_T} = {{p'}_A} + {{p'}_B}\,\,\,...{\text{(i)}} \cr & {\text{We know that,}}\,\,{{p'}_A} = {p_A}{\chi _A} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{p'}_B} = {p_B}{\chi _B} \cr} $$
Substituting the values of $${{p'}_A}$$  and $${{p'}_B}$$  in Eq. (i)
$${p_T} = {p_A}{\chi _A} + {p_B}{\chi _B}$$
Mole fraction, $${\chi _A} + {\chi _B} = 1 \Rightarrow {\chi _B} = 1 - {\chi _A}$$
On substituing in above equation,
$$\eqalign{ & {p_T} = {p_A}{\chi _A} + {p_B}\left( {1 - {\chi _A}} \right) \cr & \,\,\,\,\,\,\, = {p_A}{\chi _A} + {p_B} - {p_B}{\chi _A} \cr & \therefore \,\,{p_T} = {p_B} + {\chi _A}\left( {{p_A} - {p_B}} \right) \cr} $$

Since the new interactions $$(A-B)$$  are stronger than $$(A - A)$$  and $$(B - B)$$  interactions, the solution shows negative deviation, i.e. less vapour pressure and hence higher boiling point.

$$\eqalign{ & {\text{Sodium sulphate dissociates as}} \cr & N{a_2}S{O_4}\left( s \right) \to 2N{a^ + } + SO_4^{ - - } \cr & {\text{hence van't}}\,{\text{hoff factor}}\,i = 3 \cr & {\text{Now}}\,\,\Delta {T_f} = i\,{k_f}.m \cr & = 3 \times 1.86 \times 0.01 \cr & = 0.0558\,K \cr} $$

$$\eqalign{ & \Delta {T_f} = {K_f} \times m \cr & \,\,\,\,\,\,\,\,\,\,\, = \frac{{{K_f} \times {W_2} \times 1000}}{{{M_2} \times {W_1}}} \cr & {W_2} = \frac{{10 \times 92 \times 600}}{{1.86 \times 1000}} \cr & \,\,\,\,\,\,\,\,\, = 296.77\,g \cr} $$

$$\eqalign{ & w = 0.15\,g,\,W = 15\,g,\,\Delta {T_b} = {0.216^ \circ }C \cr & {k_b} = 2.16,\,m = ? \cr & m = \frac{{1000 \times {k_b}}}{{\Delta {T_b}}} \times \frac{w}{W} \cr & \,\,\,\,\,\, = \frac{{1000 \times 2.16 \times 0.15}}{{0.216 \times 15}} \cr & \,\,\,\,\,\, = 100 \cr} $$

When the aqueous solution of one $$molal\,KI$$   is diluted with water, concentration decreases, therefore the vapour pressure of the resulting solution increases.

$$CaC{l_2}$$  acts as a non-volatile solute and results in depression in freezing point. Thus, snow is reduced and prevents blocking of roads in the polar region.

Equimolar solutions of normal solutes in the same solvent will have the same b. pts and same f. pts.