The vapour pressure of a dilute aqueous solution of glucose is 720 mm Hg at 373 K. What is the mole fraction of the solute? (Vapour pressure of pure water at 373 K = 760 mm Hg)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1/19
Glucose is a non-volatile solute. By Raoult’s law, the relative lowering of vapour pressure of the solvent equals the mole fraction of the solute, x2.
\( \displaystyle \frac{p_1^0-p_1}{p_1^0}=x_2 \)
\( \displaystyle x_2=\frac{760\ \mathrm{mm\,Hg}-720\ \mathrm{mm\,Hg}}{760\ \mathrm{mm\,Hg}}=\frac{40}{760} \)
\( \displaystyle x_2=\frac{1}{19} \)