Question

A quantity, 10.0g, of a nonvolatile solute is dissolved in 100g of benzene (C6H6). The vapor pressure of pure benzene at 30 degrees celcius is 121.8mmHg, and that of the solution is 113.0 mmHg at the same temperature. What is the molar mass of the solute?

Answer 1

pure benzene vapoure pressure = Po = 121.8 mmHg

solution vapour pressure = Ps = 113.0 mmHg

relative lowering vapour pressure = Po - Ps / Po = 121.8 - 113 / 121.8 = 0.07225

moles of solvent benzene = 100 / 78 = 1.282 = n2

moles of solute = n1

according to raoult's law relative lowering vapour pressure = mole fraction of solute

0.07225 = n1 / n1 + 1.282

0.09263 + 0.07225 n1 = n1

0.09263 = 0.92775 n1

n1 = 0.0998

moles of solute = 0.0998

molar mass = 10 / 0.0998

                     = 100 g /mol

molar mass = 100 g/mol