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.