Solutions MCQ Questions & Answers in Physical Chemistry | Chemistry

When unripe mango is dipped in saturated salt solution then it loses water due to osmosis.

During osmosis, flow of water through a semipermeable membrane is from solution having lower concentration only.

$$\eqalign{ & \pi V = CRT \cr & \frac{{{\pi _1}}}{{{\pi _2}}} = \frac{{{C_1}{T_1}}}{{{C_2}{T_2}}} \cr & {\pi _1} = P,{\pi _2} = 2\,atm.\,{C_1} = C,{C_2} = \frac{C}{2} \cr & {T_1} = 600\,K,\,{T_2} = 700\,K \cr & \frac{P}{2} = \frac{{2 \times C \times 600}}{{C \times 700}} \cr & P = \frac{{24}}{7}. \cr} $$

$$\frac{{{p^ \circ } - {p_s}}}{{{p^ \circ }}} = \frac{{{n_2}}}{{{n_1} + {n_2}}}$$

$$\eqalign{ & i = \frac{{5.58}}{{1.86}} = 3; \cr & \left[ {Cr{{\left( {{H_2}O} \right)}_5}Cl} \right]C{l_2}.{H_2}O \to {\left[ {Cr{{\left( {{H_2}O} \right)}_5}Cl} \right]^{2 + }} + 2C{l^ - } \cr} $$

$$\eqalign{ & \pi = \frac{{{n_B}RT}}{V};{n_B} = \frac{{{W_B}}}{{{M_B}}};\pi = \frac{{{W_B}}}{{{M_B}}} \times \frac{{RT}}{V} \cr & {M_B} = \frac{{{W_B}}}{V} \times \frac{{RT}}{\pi } \cr & \,\,\,\,\,\,\,\,\, = \frac{{2 \times 0.0821 \times 300 \times 760}}{{0.3 \times 20}} \cr & \,\,\,\,\,\,\,\,\, = 6239.6\,g\,mo{l^{ - 1}} \cr} $$

Among the given options, $$CaC{l_2}$$  solution will produce maximum (three) $$ions$$  per molecule, so it will show minimum freezing point.

$$\pi = \frac{{nRT}}{V}\,\,\,or\,\,T = \frac{{\pi V}}{{nR}}$$
$$10\% \left( {w/v} \right)$$   aqueous solution of glucose means $$10\,g$$  glucose is present in $$100\,mL$$   of $${H_2}O$$
$$\eqalign{ & = \frac{{14 \times 100 \times {{10}^{ - 3}} \times 180}}{{0.082 \times 10}} \cr & = 307.3\,K \cr} $$

Equimolal solutions show same colligative properties i.e., equal elevation in boiling point and equal depression in freezing point.

$$\eqalign{ & {\text{Density}} = 1.17\,g/cc \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1170\,g/L \cr & {\text{Molarity of solution}} \cr & = \frac{{{\text{Strength in}}\,g/L}}{{{\text{Molecular weight}}}} \cr & = \frac{{1170}}{{36.5}}M \cr & = 32.05\,M \cr} $$