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Question: Molar conductivity of \(0.15M\) solution of KCl at 298 K, if its conductivity is \(0.0152\) S \(cm^{...

Molar conductivity of 0.15M0.15M solution of KCl at 298 K, if its conductivity is 0.01520.0152 S cm−1cm^{- 1} will be

A

124Ω−1cm2mol−1124\Omega^{- 1}cm^{2}mol^{- 1}

B

204Ω−1cm2mol−1204\Omega^{- 1}cm^{2}mol^{- 1}

C

101Ω−1cm2mol−1101\Omega^{- 1}cm^{2}mol^{- 1}

D

300Ω−1cm2mol−1300\Omega^{- 1}cm^{2}mol^{- 1}

Answer

101Ω−1cm2mol−1101\Omega^{- 1}cm^{2}mol^{- 1}

Explanation

Solution

Δm=k×1000M=1.52×10−2×10000.15\Delta_{m} = \frac{k \times 1000}{M} = \frac{1.52 \times 10^{- 2} \times 1000}{0.15}

=101Ω−1cm2mol−1= 101\Omega^{- 1}cm^{2}mol^{- 1}