Question

Vapour pressure of water at 293 Kis 17.535 mm Hg. Calculate the vapour pressure of water at 293 Kwhen 25 g of glucose is dissolved in 450 g of water.

Answer 1

Vapour pressure of water, = 17.535 mm of Hg

Mass of glucose, w₂ = 25 g

Mass of water, w₁ = 450 g

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

Then, number of moles of glucose,

= 0.139 mol

And, number of moles of water,

= 25 mol

We know that,

⇒ 17.535 − p₁ = 0.097

⇒ p₁ = 17.44 mm of Hg

Hence, the vapour pressure of water is 17.44 mm of Hg.