When 2.55g of an unknown weak acid (HA) with a molar mass of 85.0 g/mol is disso

Question

When 2.55g of an unknown weak acid (HA) with a molar mass of 85.0 g/mol is dissolved in 250.0 g of water, the freezing point of the resulting solution is -0.258 Degrees Celsius.

Calculate Ka for the unknown weak acid.

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When I calculated it according to the steps indicated on this webpage: https://answers.yahoo.com/question/index?qid=20090325224657AAZHvM1

I got a Ka of 0.003657 which was not correct.

Can you pinpoint what is wrong?

Thank You in advance.

Explanation / Answer

moles acid = 2.55 / 85.0 =0.0300
molality = 0.0300 / 0.250 Kg = 0.120
0.255 = 0.120 x 1.86 x i
i = Vant'Hoff factor =1.14
the degree of dissociation is
alfa = 1.14 - 1 = 0.14

K = 0.120 x( 0.14 )^2/ 1 - 0.14 =0.00273

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