The vapour pressure of water at $23℃$ is 19.8 mm. 0.1 mole of glucose is dissolved in 178.2 g of water. What is the vapour pressure (in mm) of the resultant solution?

Question

Dr. Abubacker · Accepted Answer

(b)
Given $p_{s}$ =19.8 mm
$n_{A} = 0.1$
$n_{B} = \frac{178.2}{18} = 9.9\$
According to Raoult’s law
$\frac{p_{s} - p}{p_{s}} = \frac{n_{A}}{n_{A} + n_{B}}$
$\frac{19.8 - p}{19.8} = \frac{0.1}{9.9 + 0.1}$
or 198-10 p = 19.8 $	imes$0.1
10 p = 198-1.98
10 p =196.02
p = 19.602 mm

Q. The vapour pressure of water at $23℃$ is 19.8 mm. 0.1 mole of glucose is dissolved in 178.2 g of water. What is the vapour pressure (in mm) of the resultant solution?

Explanation

Reviewed by Dr. Abubacker

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