# Molecular Weight by Freezing Point Depression of Solution

A 0.573g sample of a nonvolatile solute is added to 15.6g of napthalene and the mixture warmed until melted. The freezing point of the solution was determined and found equal to 78.1C. The freezing point of naphthalene is 80.2C. How do you find out if the boiling pt and the vapor pressure increases or decreases?

How do you find the molecular weight?

Also how do I find the highest to lowest freezing points? Example-0.1m Nacl, 0.1m C12H22O11, 0.1m Ba(NO3)2

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A 0.573g sample of a nonvolatile solute is added to 15.6g of napthalene and the mixture warmed until melted. The freezing point of the solution was determined and found equal to 78.10C. The freezing point of naphthalene is 80.20C. How do you find out if the boiling pt. and the vapour pressure increases or decreases?

How do you find the molecular weight?

Solution: Depression in freezing point Tf = Tf0- Tf, where:

Tf = Freezing Point of the solvent in the solution = 78.1 0C

Tf0 = Freezing Point of the pure solvent = 80.2 0C

Therefore, Tf = Tf0- Tf = 78.1 - 80.2 = ...

#### Solution Summary

This solution contains step-by-step calculations and explanations to determine the molecular weight, boiling point, vapour pressure and the highest to lowest freezing points. All workings are shown using equations of freezing point and molarity and a graph of freezing points is included.