Question

Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.

Answer 1

The correct answer is: 17.44 mm of Hg.
Vapour pressure of water, $p^o_1 = 17.535\, mm\, of\, Hg$
Mass of glucose, $w_2 = 25 g$
Mass of water, $w_1 = 450 g$
We know that,
Molar mass of glucose $(C_6H_{12}O_6), M_2 = 6 × 12 + 12 × 1 + 6 × 16$
$= 180 g mol ^{- 1 }$
Molar mass of water, $M_1 = 18\, g mol^{ - 1}$
Then, number of moles of glucose, $n_2=\frac{25}{180gmol^{-1}}$
=0.139mol
And, number of moles of water, $n_1=\frac{450g}{18gmol^{-1}}$
= 25 mol
We know that,
$p^o_1-p_1=\frac{n_1}{n_2+n_1}$
$⇒\frac{17.535-p_1}{17.535}=\frac{0.139}{0.139+25}$
$⇒17.535-p_1=\frac{0.139 \times 17.535}{25.139}$
$⇒ 17.535 - p_1 = 0.097$
$⇒ p_1 = 17.44 mm$ of Hg
Hence, the vapour pressure of water is 17.44 mm of Hg.