Question

$18\, g$ of glucose is dissolved in $178.2\, g$ of water. The vapour pressure of the solution at $100^{\circ}C$ is (vapour pressure of pure water at $100^{\circ}C$ is $760\, mm \,Hg$)

• A. $767.6\, mm \,Hg$
• B. $760\, mm\, Hg$
• C. $752.4\, mm \,Hg$
• D. $725.4\, mm \,Hg$
• E. $745.2\, mm \,Hg$

Answer 1

Answer: (C)

Vapour pressure $\left(p^{\circ}\right)$ of pure water at $100^{\circ} C$ is

$=760\, mm\, Hg$

From relative lowering of vapour pressure

$\frac{p^{\circ}-p_{s}}{p^{\circ}} =\frac{W_{2}}{m_{2}} \times \frac{m_{1}}{W_{1}}$

$\frac{760-p_{s}}{760} =\frac{18}{180} \times \frac{18}{178.2}=\frac{1}{99}$

$760-p_{s} =\frac{760}{99}$ or $760-p_{s} =7.6$

$\Rightarrow p_{s} =752.4\, mm\, Hg$