Question

A solution containing 2 g of glucose (M = 180 g mol⁻¹) in 100 g of water is prepared at 303 K. If the vapour pressure of pure water at 303 K is 32.8 mm Hg, what would be the vapour pressure of the solution? 

Answer 1

${\displaystyle \text{Number of moles of water}=\frac{\text{Mass of water}}{\text{mol mass of water}}}$

= ${\displaystyle \frac{100}{18}}$

= 5.55 moles

${\displaystyle \text{Number of moles of glucose}=\frac{\text{Mass of glucose}}{\text{mol mass of glucose}}}$

= ${\displaystyle \frac{2}{180}}$

= 0.0111

Total moles = 5.55 + 0.011

= 5.56 mol

${\displaystyle x_{\text{Water}}=\frac{5.55}{5.56}}$

${\displaystyle {\text{P}}_{\text{solution}}={\text{x}}_{\text{water}}\times {\text{p}}_{\text{water}}}$

P_{s =} 0.99 × 32.8

= 32.74 mm of Hg