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Question: Limiting molar conductivity for some ions is given below (in S \(cm^{2}mol^{- 1}\)): \[{Na^{+} - 50...

Limiting molar conductivity for some ions is given below (in S cm2mol−1cm^{2}mol^{- 1}):

}{- 40.9,Ca^{2 +} - 119.0}$$ What will be the limiting molar conductivities $(\Delta º_{m})$ of $CaCl_{2},CH_{3}COONa$ and NaCl respectively?
A

97.65,111.0andNaCl242.8Scm2mol−197.65,111.0andNaCl242.8Scm^{2}mol^{- 1}

B

195.3,182.0and26.2Scm2mol−1195.3,182.0and26.2Scm^{2}mol^{- 1}

C

271.6,91.0and126.4Scm2mol−1271.6,91.0and126.4Scm^{2}mol^{- 1}

D

119.0,1024.5and9.2Scm2mol−1119.0,1024.5and9.2Scm^{2}mol^{- 1}

Answer

271.6,91.0and126.4Scm2mol−1271.6,91.0and126.4Scm^{2}mol^{- 1}

Explanation

Solution

ΔºmCaCl2=λºCa2++2λºCl−\Delta º_{mCaCl_{2}} = \lambda º_{Ca^{2 +}} + 2\lambda º_{Cl^{-}}

=119.0+2×76.3=271.6Scm2mol−1= 119.0 + 2 \times 76.3 = 271.6Scm^{2}mol^{- 1}

ΔºmCH3COONa=λºCH3COO+λºNa+\Delta º_{mCH_{3}COONa} = \lambda º_{CH_{3}COO} + \lambda º_{Na^{+}}

=40.9+50.1=91Scm2mol−1= 40.9 + 50.1 = 91Scm^{2}mol^{- 1}

ΔºmNaCl=λºNa++λºCl−\Delta{º_{m}}_{NaCl} = \lambda º_{Na^{+}} + \lambda º_{Cl^{-}}

=50.1+76.3=126.4Scm2mol−1= 50.1 + 76.3 = 126.4Scm^{2}mol^{- 1}