Of a non-volatile solute is dissolved in 365.0 g of water. the solute does not react with water nor dissociate in solution. assume that the resulting solution displays ideal raoult's law behaviour. at 70°c the vapour pressure of the solution is 231.16 torr. the vapour pressure of pure water at 70°c is 233.70 torr. calculate the molar mass of the solute (g/mol).

By using this formula of vapor pressure:

Pv (solu) = n Pv (water)

when we have Pv (solu) = 231.16 torr & Pv (water) = 233.7 torr

from this formula, we can get n (mole fraction of water) by substitution:

231.16 = n * 233.7

∴ n (mole fraction of water) = 0.99

so mole fraction of solution = 1 - 0.99 = 0.01

when no. of moles of water = mass weight / molar weight

= 365g / 18g/mol = 20 moles

Total moles in solution = moles of water / mole fraction of water

= 20 / 0.99 = 20.2

no. of moles of the solution = total moles in solution - moles of water

= 20.2 - 20 = 0.2 moles

when we assumed the mass weight of the solution = 16 g (missing in your question should be given)

∴ molar mass = mass weight of solute / no. of moles of solute

= 16 g / 0.2 mol = 80 g/mol