Question

What weight of glucose (mol. wt. = 180) should be added to 1700 g of water at 20°C to lower its vapour pressure by 0.001 mm? The vapour pressure of water at 20°C is 17 mm Hg.

Answer 1

Given: Molecular weight of glucose (M₂) = 180 g/mol

Mass of water (solvent) (w₁) = 1700 g = 1.7 kg

Decrease in vapour pressure (ΔP) = 0.001 mm Hg

Vapour pressure of pure water (P°) = 17 mm Hg

M₁ ​= 18 g/mol (for water)

Let the mass of glucose = w₂ g

By using the formula of Relative lowering of vapour pressure

`(P^circ - P)/P^circ = n_2/n_1`

`(P^circ - P)/P^circ = (w_2//M_2)/(w_1//M_2)`

Since it’s a dilute solution, we can write

`(Delta P)/P^circ = (w_2 * M_1)/(w_1 * M_2)`

∴ `0.001/17 = (w_2 xx 18)/(1700 xx 180)`

⇒ 5.882 × 10⁻⁵ 

= `(18 w_2)/(306000)`

⇒ 18 w₂ = 306000 × 5.882 × 10⁻⁵

18 w₂ = 17.99892 ≈ 18

`w_2 = 18/18`

w₂ = 1 g

∴ The required weight of glucose is 1 g.