The freezing points of each of the solutions are as follows;
0.100 m K2S - - 0.558oC
21.5 g of CuCl2 in 4.50 * 102 g water - -2oC
5.5% NaNO3 by mass (in water) - - 2.6oC
What is freezing point?
The freezing point is the point at which liquid changes to solid. Let us now look at the freezing point of each solution.
a)
Since;
ΔT = K m i
K = 1.86 oC m-1
m = 0.100 m
i = 3
ΔT = 1.86 oC m-1 * 0.100 m * 3 = 0.558oC
Freezing point = 0oC - 0.558oC = - 0.558oC
b) Number of moles of CuCl2 = 21.5 g/134.45 g/mol = 0.16 moles
molality = 0.16 moles/0.45 Kg = 0.36 m
ΔT = K m i
ΔT = 1.86 oC m-1 * 0.36 m * 3 = 2oC
Freezing point = 0oC - 2 = -2oC
c) Number of moles of NaNO3 = 5.5g/85 g/mol = 0.065 moles
molality of the solution = 0.065 moles/0.0945 Kg = 0.69 m
ΔT = 1.86 oC m-1 * 0.69 m * 2 = 2.6oC
Freezing point = 0oC - 2.6oC = - 2.6oC
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