Answer to Question #269987 in General Chemistry for jacquie

Question

Calculate the vapor pressure lowering for a solution prepared at 25ºC by adding 50.0 mL of the nonvolatile glycerine (C₃H₈O₃, 92.1 g/mol, density 1.26 g/mL) to 500 mL of water (H₂O, 18.0 g/mol, density 1.00 g/mL). The vapor pressure of pure water at 25ºC is 3.17 kPa.

Answer 1

Partial pressure by solvent (when a non volatile solute is used) = mole fraction of solvent * vapour pressure of pure solventmolefractionofsolvent∗vapourpressureofpuresolvent

Therefore, no. of moles of water in 500mL = 500/18

= 27.78

no. of moles of glycerine =1.26*50/92.1

= 0.684

total moles in solution = 28.464

vapour pressure of solution = $27.78*23.8/28.64$

= 23.085 Torr