Solutions MCQ Questions & Answers in Physical Chemistry | Chemistry

Moles of solute $$ = \frac{{120}}{{60}} = 2\,moles$$
Mass of solution $$ = 1000 + 120 = 1120\,g$$
Volume of solution $$ = \frac{M}{d} = \frac{{1120}}{{1.15}} = 974\,mL$$
Molarity
$$\eqalign{ & = \frac{{{\text{Moles}}}}{{{\text{Volume of solution }}(mL)}} \times 1000 \cr & = \frac{{2 \times 1000}}{{974}} \cr & = 2.05\,M \cr} $$

$$\eqalign{ & {\text{Molarity}} = \frac{{{\text{moles of solute}}}}{{{\text{moles of solute}}\left( \ell \right)}} \cr & {\text{Mass of solution = 1000 + 120 = 1120}} \cr & {\text{d = }}\frac{M}{\upsilon };\upsilon = \frac{M}{d} = \frac{{1120}}{{1.15}}mL = \frac{{120 \times 1.15}}{{60 \times 1120}} \times 1000 = 2.05M \cr} $$

$$\eqalign{ & M = \frac{{{W_B}RT}}{{\pi V}} \cr & \,\,\,\,\,\, = \frac{{20 \times 0.0821 \times 288 \times 1000}}{{\left( {\frac{{600}}{{760}}} \right) \times 500}} \cr & \,\,\,\,\,\, = 1198 \cr} $$

Acetic acid contain carboxylic group $$– COOH$$   which can form $$H$$-bonding so acetic acid dimerises.

$$\pi \propto iCT;iCT$$    is maximum for $$0.03\,M\,NaCl$$    at $${50^ \circ }C.$$

When $$A$$ and $$B$$ form an ideal solution, $$\Delta {H_{mix}} = 0$$

The value of $${K_f}$$  depends upon the nature of the solvent. Thus, two different solutions of sucrose of same molality prepared in different solvents will have the different depression in freezing point.

NOTE: At the freezing point liquid and solid remain in equilibrium. If a solution of a non-volatile solute is cooled to a temperature below the freezing point of solution, some of liquid solvent will separate as a solid solvent and thus the concentration of solution will increase.

As $$A$$  is more volatile, vapour phase would be richer in $$A.$$  Hence $${\left( {{x_A}} \right)_{{\text{liquid}}}} < {\left( {{x_A}} \right)_{{\text{vapour}}}}$$

When a non-volalite solute $$(NaCl)$$  is dissolved in a liquid, its vapour pressure decreases.