Question

van't Hoff factor for a dilute aqueous solution of HCN is 1.00002 .
The percent degree of dissociation of the acid is​

Answer 1

Answer:

Explanation:

At infinite dilution, a weak acid dissociates completely. Acetic acid is a weak acid and therefore its dissociation increases with increase in dilution.

Above a certain dilution value, acetic acid dissociates completely

so vantoffs factor= 1+nD (D= degree of dissociation)( here n=1)

               1.00002=1+D

So D= 2x10^-5

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Answer 2

Degree of Dissociation : The dissociation degree is the fraction of original solute molecules that have dissociated. It refers to the amount of solute dissociated into ions or radicals per mole.

Van't Hoff factor : The van’t Hoff factor  is the number of moles of particles formed in solution per mole of solute. It is a property of the solute and does not depend on concentration for an ideal solution.

Given : Van't Hoff factor of HCN is 1.00002

To Find : Degree of dissociation(D) of acid.

Solution :  van't hoff  factor =  1 + n × D          ∴ ( n=1)

                                1.00002 = 1 + D

So,  D = $D = 2 * 10^{-5}$

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