Answer to Question #332514 in General Chemistry for Thel

Question

A solution that contains 12.6 g of a nonvolatile non-dissociating solute in 400 g of benzene freezes at 3.6°C. The normal freezing point of benzene is 5.5°C. What is the molar mass of the solute? (Kf for benzene = 4.96°C kg/mole)

Answer 1

The problem asks to solve for the molar mass of the solute. First, we need to identify the quantities to be used. These are

m = molality

mass = mass of solute

n = number of moles of solute

MM = molar mass of solute

kilograms of solvent

molal freezing-point depression constant

freezing-point depression

freezing point of the pure solvent

freezing point of the solution

Second, we need to identify the given values. These are

mass = 12.6 g

400 g = 0.400 kg

4.96 °C • kg/mol

Tf0 = 5.5°C

Tf = 3.6°C

Third, we need to find the freezing-point depression of the solution. The freezing-point depression is

deltaT = 5.5-3.6 = 1.9

Fourth, we need to find the molality of the solution. The molality is

m = deltaT / Kf

m = 1.9 / 4.96 = 0.38306 mol/kg

Fifth, we need to find the number of moles of solute. The number of moles of solute is

n = m(kg_solu)

n = 0.3836 x 0.4 = 0.15322 mol

Finally, we can now solve for the molar mass of solute. Therefore, the molar mass of solute is

M = m / n = 12.6 / 0.15322 = 82.2 g/mol