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Question: Molar conductivity of a solution is \(1.26 \times 10^{2}\Omega^{- 1}cm^{2}mol^{- 1}.\) Its molarity ...

Molar conductivity of a solution is 1.26×102Ω−1cm2mol−1.1.26 \times 10^{2}\Omega^{- 1}cm^{2}mol^{- 1}. Its molarity is 0.01. Its specific conductivity will be

A

1.26×10−51.26 \times 10^{- 5}

B

1.26×10−31.26 \times 10^{- 3}

C

1.26×10−41.26 \times 10^{- 4}

D

0.00630.0063

Answer

1.26×10−31.26 \times 10^{- 3}

Explanation

Solution

Λm=κ×1000Molarity\Lambda_{m} = \frac{\kappa \times 1000}{\text{Molarity}} or κ=Λm×Molarity1000\kappa = \frac{\Lambda_{m} \times \text{Molarity}}{1000}

=1.26×10−2×0.011000=1.26×10−3.= \frac{1.26 \times 10^{- 2} \times 0.01}{1000} = 1.26 \times 10^{- 3}.