The van't Hoff factor of a
0.005M aqueous solution of
KCl is 1.95. The degree of
ionization of KCl is
Answers
Answered by
7
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
Answered by
5
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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