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 = [tex]\frac{moles of solute}{mass of solvent in kilograms}[/tex]
putting the value in the equation:
molality= [tex]\frac{1.31}{0.53}[/tex]
= 2.47 M
Putting the values in freezing point equation
ΔT = 1.31 x 29.8 x 2.47
ΔT = 73.606 degrees
Given:
The molality (m),
= [tex]\frac{Moles \ of \ solute}{Mass \ of \ solvent}[/tex]
= [tex]\frac{1.31}{0.53}[/tex]
= [tex]2.47 \ M[/tex]
hence,
The freezing point of mixture will be:
→ [tex]\Delta T = iKf.m[/tex]
By substituting the values, we get
[tex]= 1.31\times 29.8\times 2.47[/tex]
[tex]= 73.606^{\circ}[/tex]
Thus the above solution is right.
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