Solutions MCQ Questions & Answers in Physical Chemistry | Chemistry

Key Idea Vapour pressure depends upon the surface area of the solution. Larger the surface area, higher is the vapour pressure.
Addition of solute decreases the vapour pressure as some sites of the surface are occupied by solute particles, resulting in decreased surface area. However, addition of solvent, i.e. dilution, increases the surface area of the liquid surface, thus results in increased vapour pressure.
Hence, addition of water to the aqueous solution of $$\left( {1\,molal} \right)\,KI,$$   results in increased vapour pressure.

Being isotonic, molar concentration of cane sugar = molar concentration of substance $$X$$
$$\frac{5}{{342}} \times \frac{{1000}}{{100}} = \frac{1}{x} \times \frac{{1000}}{{100}}$$     ( let the molecular weight of $$X = x$$   )
$$x = \frac{{342}}{5} = 68.4$$

$$\eqalign{ & {\text{According to Raoult}}'{\text{s}}\,\,{\text{law,}} \cr & \frac{{{p^ \circ } - p}}{{{p^ \circ }}} = \frac{{w \times M}}{{m \times W}} \cr} $$
$$\frac{{640 - 600}}{{640}} = \frac{{2.175 \times 78}}{{m \times 39.08}}$$     $$\left( {M\,{\text{for}}\,{C_6}{H_6} = 78} \right)$$
$$\eqalign{ & m = \frac{{2.175 \times 78 \times 640}}{{40 \times 39.08}} \cr & \,\,\,\,\,\, = 69.45 \approx 69.4 \cr} $$

$$\eqalign{ & \Delta {T_f} = {K_f}m \cr & {\text{where }}m = {\text{molality}} \cr & 273 - 268 = 1.86 \times \frac{w}{{M \times v}} \cr & 5 = 1.86 \times \frac{w}{{32 \times 10}} \cr & w = \frac{{5 \times 32 \times 10}}{{1.86}} \cr & = 860.2 \approx 868.06\,g \cr} $$

Molality of solution = $$mole$$  of solute per $$kg$$  of solvent
So, $$1$$ $$m=1$$  $$mole$$  of solute per $$1000 g$$  of solvent
Hence,
$$moles$$  of solute in $$1$$ $$m$$  aqueous solution $$= 1$$
$$moles$$  of solvent in $$1$$ $$m$$  aqueous solution
$$\eqalign{ & = \frac{{1000}}{{18}} \cr & = 55.55 \cr} $$
$$Mole$$  fraction of solute in $$1$$ $$m$$  solution
$$\eqalign{ & = \frac{1}{{1 + 55.55}} \cr & = \frac{1}{{56.55}} \cr & = 0.0176 \approx 0.018 \cr} $$

$$\eqalign{ & {\text{Moles of solute}} = \frac{{28}}{{28}} = 1; \cr & {\text{moles of water}} = \frac{{100 - 28}}{{18}} = 4 \cr & V.P\,{\text{of solution}} \cr & = P_{{H_2}O}^ \circ \times {X_{{H_2}O}} \cr & = 760 \times \frac{4}{5} = 608\,torr \cr} $$

$$\eqalign{ & KCl \to {K^ + } + C{l^ - },{\text{thus,}}\,i = 2 \cr & NaCl \to N{a^ + } + C{l^ - },{\text{thus,}}\,i = 2 \cr & {K_2}S{O_4} \to 2{K^ + } + SO_4^{2 - }\,{\text{thus,}}\,i = 3 \cr} $$

Osmotic pressure is an example of colligative property because its value depends only on the number of moles of solute, not on their chemical nature.

For the same concentration of different solvents any colligative property $$ \propto {\text{i}}$$
For $$NaCl,\,i = 2$$
Sugar solution, $$i = 1$$
$$BaC{l_2},\,i = 3;\,\,\,FeC{l_3},\,i = 4$$
Thus, for sugar solution depression in freezing point is minimum i.e., highest freezing point.

$$\eqalign{ & {\text{Moles of urea}} = \frac{5}{{60}}; \cr & {\text{moles of fructose}} = \frac{5}{{180}}; \cr & {C_{urea}} = \frac{{0.08 \times 1000}}{{100}} \Rightarrow 0.833 \cr & {\text{moles of sucrose}} = \frac{5}{{342}}; \cr & {C_{{\text{sucrose}}}} = 0.0146 \times \frac{{1000}}{{100}} = 0.146 \cr & moles{\text{ of }}KCl\,{\text{(effective)}} \cr & {\text{ = 2}} \times \frac{5}{{74.5}} = \frac{5}{{37.25}}; \cr & \frac{{0.134 \times 1000}}{{100}} = 1.34 \cr & \pi \propto C \cr & {\text{this order is}}\,\,{\pi _4} > {\pi _1} > {\pi _2} > {\pi _3} \cr} $$