Question

An unknown nonelectrolyte and nonvolatile compound is dissolved in enough pure water to make 348ml of solution.The vapor pressure above the solution drops from 198.7mbar to 196.7mbar. When 100g of the compound is dissolved at constant temperature.Determine the molar mass of the unknown compound. Assume no change in volume occurs upon dissolution.

The answer is 509 g/mol Why?

Answer 1

Let the molar mass of the solution be M g/mol; we start with 100 g of the compound. Therefore, moles of compound taken = (100 g)*(1 mol/M g) = 100/M mol.

Assume the density of water to be 1 g/mL so that the mass of water taken = (348 mL)*(1 g/1 mL) = 348 g.

Molar mass of water = 18 g/mol; therefore, moles of water taken = (348 g)*(1 mol/18 g) = 19.333 mol.

Mole fraction of solvent (water) in the mixture = 19.333/(19.333 + 100/M)

As per Raoult’s law,

P_soln = (mole fraction of solvent)*P⁰_solvent where P_soln is the vapour pressure of the solution and P⁰_solvent is the vapour pressure of the pure solvent. Plug in values to obtain

(196.7 mbar) = [19.333/(19.333 + 100/M)]*(198.7 mbar)

===> 0.9899 = 19.333/(19.333 + 100/M)

===> 0.9899*(19.333 + 100/M) = 19.333

===> 19.1377 + 98.99/M = 19.333

===> 98.99/M = 19.333 – 19.1377 = 0.1953

===> M = 98.99/0.1953 = 506.86 ≈ 507

The molar mass of the solute is 507 g/mol; this is slightly off from the given answer possibly because of different rounding off of digits (ans).