Question

Vapour pressure of water at 293k is17.535mmHg. Calculate the vapour pressure of water at 293k when 25g of glucose is dissolved in 450 g of water

Answer 1

Glucose formula = (C6H12O6)

The vapour pressure of water, Po1 = 17.535 mm Hg

Given mass of glucose = 25 g

Given mass of water = 450 g

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

Molar mass of water = 18 g mol−1

Number of moles = Mass/ Molar Mass

                             = 450/18

                             = 25 mol

Using the same formula again we get

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

As per Raoult’s law -

17.535-p1/17.535 = 0.139/0.139+25

= 17.44 mm of Hg