Question

The vapour pressure of water at $20$ is 17.5 mm Hg. If 18 g of glucose $C_6{H}_{12}{O}_6$ is added to 178.2 g water $20$ , the vapour pressure of the resulting solution will be:

• A. 17.675 mm Hg
• B. 15.750 mm Hg
• C. 16.500 mm Hg
• D. 17.325 mm Hg

Answer 1

Answer: (D)

17.325 mm Hg

No. of moles of glucose=$\frac{18}{180}=0.1$

No. of moles of water=$\frac{178.2}{18}=9.9$

Mole fraction of glucose=$\frac{0.1}{4.9+0.1}=\frac{0.1}{10}=0.01$

Using Raoult's law, as solute is non-volatile,

$x_{\beta }=\frac{P_A^o-P_A}{P_A^o}$

$P_A^o=$vapour pressure of pure water

$P_A=$vapour pressure of the solution

$x_{\beta }=$mole fraction of glucose

$⟹0.01=\frac{17.5-P_A}{17.5}⟹P_A=17.5-0.175=17.325$

The vapor pressure of the mixture=17.325 mm Hg.

Hence, the correct option is $D$