Question

The van't Hoff factor of a
0.005M aqueous solution of
KCl is 1.95. The degree of
ionization of KCl is

Answer 1

Solution:

i = 1 + (n-1)∝

where,

i => Van't Hoff Factor

n => Number of particles

∝ => Degree of Dissociation/Ionization

i = 1.95

n = 2

n i.e. number of particles of KCl will be 2 because we know that KCl gets ionized by two ions i.e. K⁺ and Cl⁻

i = 1 + (n-1)∝

1.95 = 1 + (2 - 1)∝

1.95 = 1 + ∝

∝ = 1.95 - 1

∝ = 0.95

The degree of ionization of KCl is 0.95

Answer 2

${\boxed{\underline{\tt{ \orange{Required \: \: answer:-}}}}}$

i = 1 + (n-1)∝

where,❁

i => Van't Hoff Factor

n => Number of particles

∝ => Degree of Dissociation/Ionization☄

i = 1.95

n = 2 ✔

n i.e. number of particles of KCl will be 2 because we know that KCl gets ionized by two ions i.e. K⁺ and Cl⁻❀

▸i = 1 + (n-1)∝

▸1.95 = 1 + (2 - 1)∝

▸1.95 = 1 + ∝

▸∝ = 1.95 - 1

▸∝ = 0.95

The degree of ionization of KCl is 0.95

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