Calculate the freezing point (kf = 1.86 °c/m) of a 2.1m aqueous solution of kcl.

7.8 °c

3.9 °c

-7.8 °c

-3.9 °c

The freezing point of the solution is - 3.90 °C.

Explanation:

ΔTf = (Kf)(m),

where, ΔTf is the depression in the freezing point.

Kf is the molal freezing point depression constant of water = - 1.86 °C/m,

m is the molality of the solution (m = 2.1 m).

∴ ΔTf = (Kf)(m) = (-1.86 °C/m)(2.1 m) = - 3.90 °C.

∵ The freezing point if water is 0.0 °C and it is depressed by - 3.90 °C.

∴ The freezing point of the solution is - 3.90 °C.

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