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3 years ago

10

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)?

1 answer:

Allisa [31]3 years ago

6 0

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 = $\frac{moles of solute}{mass of solvent in kilograms}$

putting the value in the equation:

molality= $\frac{1.31}{0.53}$

= 2.47 M

Putting the values in freezing point equation

ΔT = 1.31 x 29.8 x 2.47

ΔT = 73.606 degrees

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Explanation:

<u>Step 1:</u> Data given

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8 0

3 years ago