Question

The vapour pressure of water at ${20}^C$ 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 resulting solution will be _

• A. 17.325 mm Hg
• B. 17.675 mm Hg
• C. 15.75 mm Hg
• D. 16.5 mm Hg

Answer 1

The correct option is A 17.325 mm Hg

Vapour pressure $=1.75mmHgat{20}^{\circ }C$
$\therefore P^{\circ }=17.5mmHg$
$\therefore P=x{p}^{\circ }$ ------ (Raoult's law)
$=\left(\frac{\frac{178.2}{18}}{\frac{18}{180}+\frac{178.2}{18}}\right)\left(17.5\right)$
$==\left(\frac{9.9}{10}\right)\left(17.5\right)$
$=17.325mmHg$