Depression of the melting point of solvents in the presence of solutes.

Question

What would be the correct steps to take in trying to solve the following question:

"In the Rast method for determining molecular weight, camphor with a melting point of 179°C is used as the solvent.If 1 mol of a compound is dissolved in 1000g of camphor, it brings the melting point of the solution down to 139°C (cryoscopic constant = 40). The following data are given:

•Weight of camphor: 1.80 g

•Weight of the unknown: 0.054 g

•Melting point of the solution: 165°C

How much is the molecular weight of the unknown?"

Answer 1

If 165° is the melting point of the pure camphor, the following equation holds:

ΔT_mp = C x m where C is the cryoscopic constant, ΔT_mp is the number of degrees the melting point of the camphor is lowered, and m is the molality of the solution.

Substitute the given values for ΔT_mp and C and solve for m.

m = ΔT_mp / C = (165 - 139) / 40 = 0.65 molal.

molality is the number of moles of solute dissolved in 1000 g of solvent (camphor).

As 0.054 g solute is dissolved in 1.80 g of solvent, this is 0.054 g solute / 1.80 g camphor x 1000 g camphor = 30 g solute in 1000 g camphor. The number of moles of solute is given by 30 g / MM solute which then equal to the calculated molality 0.65 = 30 / MM where MM is the molecular mass of the solute that we are trying to determine.

So, m = 30 g / MM and solving for MM you get 30 g / 0.65 = 46.1 g / mole.

Check all math.