Question

A 0.1 molal aqueous solution of a weak acid is 30% ionized. if k f for water is 1.86c/m, the freezing point of the solution will be

Answer 1

Answer. -0.24℃

Explanation:

For weak acid i=1+0.3

As weak acid is 30% ionised.

Then ∆Tf=Kf.m

∆Tf=1.86×0.1×1.3

∆Tf=0.24℃

So Tf= -0.24℃.

Answer 2

Answer:

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