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
