Calculate the vapor pressure lowering on an aqueous solution of glycerin that contains 180g of glycerin and 1000g of water at 25c. the vapor pressure of water at this temperature is 23.8torr.

The vapor pressure of a solvent is a colligative property. That means that it depends on the number of particles of the solvent in solution and not the nature of the solute dissolved.

You have to use Raoult's law, which states that the vapor pressure of a solution is equal to the vapor pressure of the pure solvent times the molar fraction of the solvent, i.e.:

     P = P°solv * X solv

X solv = number of moles of solvent / number of moles of solution

solvent = water

solute = glycerin

number of moles of water = mass of water / molar mass of water

number of moles of water = 1000 g / 18.02 g/mol = 55.49 mol

number of moles of glycerine = mass of glycerin / molar mass of glycerin

number of moles of glycerin = 180g / 92.09 g/mol = 1.955 mol

X solv = 55.49 mol / (55.49 mol + 1.955 mol) = 0.966

P = 23.8 torr * 0.966 = 22.99 torr = 23.0 torr

=> Vapor pressure lowering = 23.8 torr - 23.0 torr = 0.8 torr

0.8 torr.

Okay this will take 1 hour okay dont forget me.

Idon’t know sorry i don’t know the answer i’m s sorry