If the molar conductivity ( _m ) of a 0.050 mol L^-1 solution of a monobasic weak acid is 90 S cm^2 mol^-1 , its extent (degree) of dissociation will - Chemistry MCQ Chemistry | ExamTest.live

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If the molar conductivity (Λ_{m}) of a 0.050 mol L^{- 1} solution of a monobasic weak acid is 90 S cm^{2} mol^{- 1}, its extent (degree) of dissociation will be[Assume λ_{+}^{\circ} = 349.6 S cm^{2} mol^{- 1} and λ_{-}^{\circ} = 50.4 S cm^{2} mol^{- 1}]

A $0.115$
B $0.225$
C $0.125$
D $0.215$

Solution & Step-by-step Explanation

1. Calculate the molar conductivity at infinite dilution (Λ_{m}^{\circ}) using Kohlrausch's law: Λ_{m}^{\circ} = λ_{+}^{\circ} + λ_{-}^{\circ} = 349.6 + 50.4 = 400 S cm^{2} mol^{- 1} .2. Calculate the degree of dissociation (α): α = \frac{Λ _{m}}{Λ _{m}^{\circ}} = \frac{90}{400} = \frac{9}{40} = 0.225 .

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If the molar conductivity (Λ_{m}) of a 0.050 mol L^{- 1} solution of a monobasic weak acid is 90 S cm^{2} mol^{- 1}, its extent (degree) of dissociation will be[Assume λ_{+}^{\circ} = 349.6 S cm^{2} mol^{- 1} and λ_{-}^{\circ} = 50.4 S cm^{2} mol^{- 1}]

0.115

0.225

0.125

0.215

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Electrochemistry Molar Conductivity Degree of Dissociation

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If the molar conductivity ( $Λ_{m}$ ) of a $0.050 mol L^{- 1}$ solution of a monobasic weak acid is $90 S cm^{2} mol^{- 1}$ , its extent (degree) of dissociation will be[Assume $λ_{+}^{\circ} = 349.6 S cm^{2} mol^{- 1}$ and $λ_{-}^{\circ} = 50.4 S cm^{2} mol^{- 1}$ ]

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If the molar conductivity (Λm​) of a 0.050mol L−1 solution of a monobasic weak acid is 90S cm2 mol−1, its extent (degree) of dissociation will be[Assume λ+∘​=349.6S cm2 mol−1 and λ−∘​=50.4S cm2 mol−1 ]

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Discussion

If the molar conductivity ( $Λ_{m}$ ) of a $0.050 mol L^{- 1}$ solution of a monobasic weak acid is $90 S cm^{2} mol^{- 1}$ , its extent (degree) of dissociation will be[Assume $λ_{+}^{\circ} = 349.6 S cm^{2} mol^{- 1}$ and $λ_{-}^{\circ} = 50.4 S cm^{2} mol^{- 1}$ ]