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

Vapour pressure of water, p⁰ = 17.535 mm of Hg

Mass of glucose, w₂ = 25 g

Mass of water, w₁ = 450 g

P_s = ?

We know that,

Molar mass of glucose (C₆H₁₂O₆), M₂ = 6 × 12 + 12 × 1 + 6 × 16

= 180 g mol⁻¹

Molar mass of water, M₁ = 18 g mol⁻¹

Applying Raoult's law,

${\displaystyle \frac{{\text{p}}^0-{\text{p}}_{\text{s}}}{{\text{p}}^0}=\frac{{\text{n}}_2}{{\text{n}}_1+{\text{n}}_2}}$,

${\displaystyle \frac{{\text{p}}^0-{\text{p}}_{\text{s}}}{{\text{p}}^0}=\frac{{\text{n}}_2}{{\text{n}}_1}=\frac{\frac{{\text{w}}_2}{{\text{M}}_2}}{\frac{{\text{w}}_1}{{\text{M}}_1}}}$  ...(∵ n₂ << n₁)

or, ${\displaystyle 1-\frac{{\text{p}}_{\text{s}}}{{\text{p}}^0}=\frac{{\text{w}}_2{\text{M}}_1}{{\text{w}}_1{\text{M}}_2}}$

Substituting the given value, we get

${\displaystyle 1-\frac{{\text{p}}_{\text{s}}}{17.535}=\frac{25\times 18}{450\times 180}}$ or, ${\displaystyle 1-\frac{{\text{p}}_{\text{s}}}{17.535}=\frac{1}{180}}$

${\displaystyle 1-\frac{1}{180}=\frac{{\text{p}}_{\text{s}}}{17.535}}$ or, ${\displaystyle \frac{179}{180}=\frac{{\text{p}}_{\text{s}}}{17.535}}$

or, p_s = ${\displaystyle 17.535\times \frac{179}{180}}$ = 17.437 mm Hg