Question

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

• A. 20.4
• B. 10.3
• C. 5.4
• D. 7.4

Answer 1

Answer: (A)

Vapour pressure of pure solvent

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

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

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

$=23.8-23.324$

$=0.476\, mm$ 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;

$M_{A} =18,\, w_{A}=100$

$w_{B} =M_{B}=180$

$w_{B} =\frac{\Delta p \times M_{B} \times w_{A}}{p^{\circ} \times M_{A}}$

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

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

$=20.00$