Answer to Question #112001 in Physical Chemistry for DEEPAK SINGH

Question

vapor pressure of water at 293k is 17.535 mm hg.calculate the vapor pressure of water at 293k when 25g of glucose is dissolved in 450g of water.

Answer 1

(P₀ - P₁)/P₀=n(glucose )/(n(glucose)+n(H2O))

P₀=17.535 mm Hg

n(glucose) = 25/180=0.1389 mol

n(H2O)=450/18=25 mol

P₁=P₀ - (n(glucose)*P₀)/(n(glucose)+n(H2O)) =17.535 - (0.1389*17.535)/(0.1389+25)=17.438 mm Hg=2,325 kPa

Answer: 17.438 mm Hg or 2,325 kPa