NEET 2025 — Chemistry solution

NEET2025Chemistry

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
0.125
C
0.225
D
0.215

Solution

$Λ_{m}^{\circ} = Λ_{+}^{\circ} + Λ_{-}^{\circ} = 349.6 + 50.4 = 400$ S cm $^{2}$ mol $^{- 1}$ .

Degree of dissociation: $α = \frac{Λ _{m}}{Λ _{m}^{\circ}} = \frac{90}{400} = 0.225$ .

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