Question

A solution of a nonvolatile solute in water freezes at – 0.30°C. The vapour pressure of pure water at 298 K is 23.51 mm Hg and $K_f$ for water is 1.86 K kg mol⁻¹. Calculate the vapour pressure of this solution at 298 K.

Answer 1

Hello Student,

Please find the answer to your question

∆Tf = Kf.m

Po – p/po = moles of solute/moles of solvent

Depression in freezing point, ∆Tf = Kf m

∴ m ∆Tf/Kf = 0.30/1.86 = 0.161

According to Raoult’s law

Po – p/po = No. of moles of solute/No. of moles of solvent

23.51 – p/23.51 = 0.161/1000/ 18 = 0.161 *18/1000

(∵ No. of moles of H2O = 1000/18)

On usual calculations,

23.51 –p/23.51 = 0020898

P = 23.51 – 23.51 * 0.0020898 = 23.51 – 068

p = 23.44 mm Hg