A sample of mass 6.814 grams is added to another sample weighing 0.08753 grams.
1 answer:
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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