171.
Vapour pressure of pure water at $$298\,K$$ is $$23.8\,\,mm\,\,Hg.$$ $$50\,g$$ of urea is dissolved in $$850\,g$$ of water. The vapour pressure of water for this solution and its relative lowering are respectively
172.
When a gas is bubbled through water at $$298\,K,$$ a very dilute solution of gas is obtained. Henry's law constant for the gas is $$100\,kbar.$$ If gas exerts a pressure of $$1\,bar,$$ the number of moles of gas dissolved in 1 litre of water is
173.
If the elevation in boiling point of a solution of $$10\,g$$ of solute $$\left( {mol.\,wt. = 100} \right)$$ in $$100\,g$$ of water is $$\Delta {T_b},$$ the ebullioscopic constant of water is
174.
The vapour pressure of water at $${20^ \circ }C$$ is $$17.5\,mm\,Hg.$$ If $$18 g$$
of glucose $$\left( {{C_6}{H_{12}}{O_6}} \right)$$ is added to $$178.2 g$$ of water at $${20^ \circ }C,$$ the vapour pressure of the resulting solution will be
NOTE : On addition of glucose to water, vapour pressure of water will decrease. The vapour pressure of a solution of glucose in water can be calculated using the relation
$$\eqalign{
& \frac{{{P^ \circ } - {P_s}}}{{{P_s}}} = \frac{{{\text{Moles of glucose in solution}}}}{{{\text{moles of water in solution}}}} \cr
& {\text{or}}\,\frac{{17.5 - {P_s}}}{{{P_s}}} = \frac{{\frac{{18}}{{180}}}}{{\frac{{178.2}}{{18}}}}\,\,\,\,\,\,\,\left[ {\because \,\,{P^ \circ } = 17.5} \right] \cr
& {\text{or}}\,17.5 - {P_s} = \frac{{0.1 \times {P_s}}}{{9.9}}\,or\,{P_s} = 17.325\,mm\,Hg. \cr
& {\text{Hence (A) is correct answer}}{\text{.}} \cr} $$
175.
For preparing $$0.1N$$ solution of a compound from its impure sample ( percentage purity of which is known ) weight of the substance required will be
More than theoretical weight since impurity will not contribute.
176.
$${K_H}$$ values for $$A{r_{\left( g \right)}},C{O_{2\left( g \right)}},HCH{O_{\left( g \right)}}$$ and $$C{H_{4\left( g \right)}}$$ are $$40.39,1.67,1.83 \times {10^{ - 5}}$$ and $$0.413$$ respectively. Arrange these gases in the order of their increasing solubility.
Higher the value of $${K_H},$$ the lower is the solubility of gas in the liquid.
Thus, the order of increasing solubility is
$$\mathop {}\limits_{{K_H}\,\,{\text{value}}\,:} \,\,\,\,\,\mathop {Ar}\limits_{40.39} < \mathop {C{O_2}}\limits_{1.67} < \mathop {C{H_4}}\limits_{0.413} < \mathop {HCHO}\limits_{1.83 \times {{10}^{ - 5}}} $$
177.
Benzene freezes at $${5.50^ \circ }C.$$ If the freezing point of $$2.48\,g$$ of phosphorous in $$100\,g$$ benzene is $${4.48^ \circ }C,$$ the atomicity of phosphorus in benzene is $$\left( {{K_f}{\text{(benzene)}} = 5.12\,K\,kg\,mo{l^{ - 1}}} \right):$$
178.
$$2\,g$$ of sugar is added to one litre of water to give sugar solution. What is the effect of addition of sugar on the boiling point and freezing point of water ?
A
Both boiling point and freezing point increase.
B
Both boiling point and freezing point decrease.
C
Boiling point increases and freezing point decreases.
D
Boiling point decreases and freezing point increases.
Answer :
Boiling point increases and freezing point decreases.
When a non-volatile solute is added to water there is elevation in boiling point and depression in freezing point.
179.
Vapour pressure of pure benzene is $$119\,torr$$ and that of toluene is $$37.0\,torr$$ at the same temperature. $$Mole$$ fraction of toluene in vapour phase which is in equilibrium with a solution of benzene and toluene having a mole fraction of toluene $$0.50,$$ will be :
180.
$$0.5\,molal$$ aqueous solution of a weak acid $$(HX)$$ is $$20\% \,ionised.$$ If $${K_f}$$ for water is $$1.86\,K\,kg\,mo{l^{ - 1}},$$ the lowering in freezing point of the solution is