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.

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.

Solution:

Given that

The vapour pressure of water, P^o₁ = 17.535 mm Hg

Given mass of glucose = 25 g

Given mass of water = 450 g

Calculate

Molar mass of glucose (C₆H₁₂O₆) = 6 × 12 + 12 × 1 + 6 × 16 = 180 g mol−1

Molar mass of water = 18 g mol⁻¹

 

Plug the values we ge

t

Number of moles of water  = 450/18  = 25 mol

Use same formula again we get

Number of moles of glucose = 25/180 = 0.139 mol

Formula of Raoult’s law

Plug the values in this formula we get 

Add and cross multiply it we get

(17.535 − p1)(1.64) = 0.097 × 17.535

After calculation we get

p₁ = 17.44 mm of Hg

So that our Solution, the vapour pressure of water =17.44 mm Hg.