Question

How many grams of glucose are required to prepare an aqueous solution of glucose having a vapour pressure of $23.324mmHg$ at $25^{\circ }C$ in $100g$ of water? The vapour pressure of pure water at $25^{\circ }C$ is $23.8mmHg$. (Molar mass of glucose $=180gmo{l}^{-1}$ )

• A. 20.4
• B. 10.3
• C. 5.4
• D. 7.4

Answer 1

Answer: (A)

Vapour pressure of pure solvent

$=p^{\circ }=23.8mm$ of $Hg$

Vapour pressure of solution $(p)=23.324mm$ of $Hg$

Thus, difference in vapour pressure $(\Delta p)$

$=23.8-23.324$

$=0.476mm$ of $Hg$

$\because \frac{\Delta p}{p^{\circ }}=\frac{w_B}{M_B}\times \frac{M_A}{w_A}$ (for dilute solution)

where, $M_A$ and $w_A$ are molar and actual masses of solvent respectively.

$M_B$ and $w_B$ are molar and actual masses of solute respectively.

Given;

$M_A=18,w_A=100$

$w_B=M_B=180$

$w_B=\frac{\Delta p\times M_B\times w_A}{p^{\circ }\times M_A}$

$=\frac{0.02\times 180\times 100}{18}$

$\left[\frac{\Delta p}{p^{\circ }}=\frac{0.476}{23.8}=0.02\right]$

$=20.00$