What is the freezing point of a solution made with 1.31 mol of CHCl3 in 530.0 g of CCl4 (Kf =29.8 degrees C/m, Freezing point of pure solvent = -22.9 degrees C)?

Given:

The molality (m),

=

=

=

hence,

The freezing point of mixture will be:

→

By substituting the values, we get

Thus the above solution is right.

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73.606 °C is the freezing point of the solution made with with 1.31 mol of CHCl3 in 530.0 g of CCl4.

Explanation:

Data given:

number of moles of CHCl3 = 1.31 moles

mass of solvent CHCl3 = 530 grams or 0.53 kg

Kf = 29.8 degrees C/m

freezing point of pure solvent or CCl4 =  -22.9 degrees

freezing point = ?

The formula used to calculate the freezing point of the mixture is

ΔT = iKf.m

m=  molality

molality =

putting the value in the equation:

molality=

             = 2.47 M

Putting the values in freezing point equation

ΔT = 1.31 x 29.8 x 2.47

ΔT = 73.606 degrees

b. examine and describe cells