Question

100 ml of 0.1 M solution of a weak acid has molar conductivity . The osmotic pressure of the resulting solution obtained after dilution of original solution upto 1 litre at 500 K, assuming ideal solution is:

Answer 1

α=0.147
Ka=1.86⋅10−3

Explanation:

First thing first, I think that you have some incorrect data in your question.

For example, if you use the information provided by the question, the molar conductivity of the chloroacetic solution will have different value than what you list here, 362 S m2mol−1.

I assume that this value was supposed to be the molar conductivity at infinite dilution, Λ0. Moreover, I think that the value you provided is incorrect.

More specifically, this value should either be

Λ0=362⋅10−4S m2mol−1

or

Λ0=362 S cm2mol−1

With this being said, the equation that establishes a relationship between conductivity, k, and molar conductivity, Λ, looks like this

Λ=kc , where

c - the molarity of the solution.

This is where things usually get a little tricky. You need to make sure that you use the right units. For example, molar concentration must be used in moles per cubic meter, which means that you must convert the given moles per liter

0.0625molesliter⋅103liters1 m3=0.0625⋅103mol m−3

Likewise, notice that the conductivity uses cm−1, so convert it to

3.319⋅10−3Scm⋅102cm1 m=3.319⋅10−1S m−1

The molar conductivity of the solution will thus be

Λ=3.319⋅10−1S m−10.0625⋅103m−3=5.3104⋅10−3S m−2mol−1

Now, the equation that connects the degree of ionization of a weak electrolyte, α, and its molar conductivity looks like this

α=ΛΛ0

In your case, the degree of dissociation for chloroacetic acid will be - keep in mind that α must come out to be unitless!

α=5.3104⋅10−3S m−2mol−1362⋅10−4S m−2mol−1=0.147

The acid dissociation constant, Ka, can be calculated using Ostwald's dilution law, which establishes a relationship between the dissociation constant and the degree of ionization of a weak electrolyte

Ka=α21−α⋅c

In your case, the value of Ka will be - use the molarity in moles per liter

Ka=0.1472(1−0.147)2⋅0.0625 M=1.86⋅10−3M

The listed value for the acid dissociation constant of chloroacetic acid is 1.4⋅10−3M, so the result is good enough.

You can check this result by using the molar conductivity

Ka=Λ2(Λ0−Λ)⋅Λ0⋅c

This calculation will give you

Ka=1.58⋅10−3M