Question

The solution of a non-volatile solute in water freezes at -0.30°C. The vapour pressure of pure water at 298 K is 23.51 mm of Hg and Kof water is 1.86 degree/molal. Calculate
the vapour pressure of this solution at 298 K.​

Answer 1

Answer: .......

Explanation:

Answer 2

∆T = K×m

=> 0.3 = 1.86 × m

=> m = 0.161

m = (Xa×1000)/(1-Xa×Mb)

0.161 = (Xa×1000)/(1-Xa×18)

=> Xa = 0.00289

Also, ∆P/P° = Xa

0.00289 = (23.51 - Ps)/23.51

=> Ps = 23.442 mm

Hence, vapour pressure of this solution is: Ps = 23.442 mm of Hg