Question
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
A.
$$17.325\,mm\,Hg$$
B.
$$15.750\,mm\,Hg$$
C.
$$16.500\,mm\,Hg$$
D.
$$17.500\,mm\,Hg$$
Answer :
$$17.325\,mm\,Hg$$
Solution :
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
$$\eqalign{
& \frac{{{P^ \circ } - {P_s}}}{{{P_s}}} = \frac{{{\text{Moles of glucose in solution}}}}{{{\text{moles of water in solution}}}} \cr
& {\text{or}}\,\frac{{17.5 - {P_s}}}{{{P_s}}} = \frac{{\frac{{18}}{{180}}}}{{\frac{{178.2}}{{18}}}}\,\,\,\,\,\,\,\left[ {\because \,\,{P^ \circ } = 17.5} \right] \cr
& {\text{or}}\,17.5 - {P_s} = \frac{{0.1 \times {P_s}}}{{9.9}}\,or\,{P_s} = 17.325\,mm\,Hg. \cr
& {\text{Hence (A) is correct answer}}{\text{.}} \cr} $$