Question

The experimental and calculated elevation in boiling

points of an electrolyte AB in its aqueous solution at a

given concentration are 0.81 K and 0.54 K

respectively. The percentage ionization of the

electrolyte at this concentration is​

Answer 1

Answer:

50 is answer dear friend

Answer 2

The percentage of ionization of the electrolyte AB will be 50%.

Given:

The experimental elevation in boiling points of an electrolyte AB = 0.81K

The calculated elevation in boiling point= 0.54K

To Find:

The percentage ionization of the electrolyte at this concentration.

Solution:

Let us assume α is the degree of dissociation.

AB--> $A^++B^-$

Initial no. of moles of AB, cation A and anion B are 1, 0, and 0 respectively.

Moles after dissociation are 1, α, and α respectively.

Van't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass.

Van’t Hoff factor, i= Moles of solute after dissociation/ Normal moles of solute

Hence, $i= \frac{1+\alpha }{1 } = 1+\alpha$

i= Normal value / Calculated value= 0.81/0.54=1.5

1+ α= 1.5

⇒α=0.5

Hence, Percentage of ionization = 0.5x 100= 50%

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