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, p° = 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,

`(p^circ - p_s)/p^circ = n_2/(n_1 + n_2)`,

`(p^circ - p_s)/p^circ = n_2/n_1 = (w_2/M_2)/(w_1/M_1)`    ...(∵ n₂ << n₁)

or, `1 - p_s/p^circ = (w_2 M_1)/(w_1 M_2)`

Substituting the given value, we get

`1 - p_s/17.535 = (25 xx 18)/(450 xx 180)`

or, `1 - p_s/17.535 = 1/180`

`1 - 1/180 = p_s/17.535`

or, `179/180 = p_s/17.535`

or, p_s = `17.535 xx 179/180`

= 17.437 mm Hg

Answer 2

In the present case,

w = 25 g, W = 450 g, p° = 17.535 mm, p = ?

Molar mass of glucose (C₆H₁₂O₆),

M' = (6 × 12) + (12 × 1) + (6 × 16)

= 180 g mol⁻¹

Molar mass of water (M) = 18

∵ `(p^circ - p)/p^circ = (wM)/(WM')`

We have 

`(17.535 - p)/17.535 = (25 xx 18)/(450 xx 180)`

or `17.535 - p = (25 xx 18 xx 17.535)/(450 xx 180)`

= 0.0974

or p = 17.535 − 0.0974

= 17.44 mm Hg