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If $Λ_{c}$ of ${NH}_{4} OH$ is $11.5 Ω^{- 1} {cm}^{2} {mol}^{- 1}$ , its degree of dissociation would be
(Given $λ_{{NH}_{4}^{+}}^{\infty} = 73.4 Ω^{- 1} {cm}^{2} {mol}^{- 1} and λ_{OH}^{\infty} = 197.6 Ω^{- 1} {cm}^{2} {mol}^{- 1} .$ )

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0.157

0.058

0.0424

0.0848

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detailed solution

Correct option is C

degree of dissociation solution is 0.0424

$α = \frac{Λ}{Λ^{\infty}} = \frac{11.5}{(73.4 + 197.6)} = \frac{11.5}{271.0} = 0.0424$ .

When free ions carrying current are produced at a specific concentration and are dissociated from the solute fraction, this phenomenon is known as the degree of dissociation.

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If Λc of NH4OH is 11.5Ω−1cm2mol−1, its degree of dissociation would be(Given λNH4+∞=73.4Ω−1cm2mol−1 and λOH∞=197.6Ω−1cm2mol−1.)

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If $Λ_{c}$ of ${NH}_{4} OH$ is $11.5 Ω^{- 1} {cm}^{2} {mol}^{- 1}$ , its degree of dissociation would be
(Given $λ_{{NH}_{4}^{+}}^{\infty} = 73.4 Ω^{- 1} {cm}^{2} {mol}^{- 1} and λ_{OH}^{\infty} = 197.6 Ω^{- 1} {cm}^{2} {mol}^{- 1} .$ )