Question

Vapour pressure of water at $293\, K$ is $17.535\, mm \,Hg$. The vapour pressure of water at $293 \,K$ containing $25 \,g$ of glucose dissolved in $450\, g$ of water is

• A. $17.439\, mm \,Hg$
• B. $17.535\, mm \,Hg$
• C. $0.097\, mm\, Hg$
• D. $34.973\, mm \,Hg$

Answer 1

Answer: (A)

$17.439\, mm \,Hg$

Answer 2

$\frac{p^{\circ} -p_{s}}{p^{\circ}} = \frac{w_{2} \times M_{1} }{w_{1} \times M_{2}}$
$\frac{17.535 - p_{s}}{17.535} = \frac{25 \times 18 }{450 \times 180} = 5.55 \times 10^{-3} p_{s} =17.437 mm Hg$