500 mL of an aqueous solution contains 0.1 mole of the solute AB; if its specific conductance is x S $C m^{- 1}$ its molar conductance will be .. S $C m^{2}$ ${m o l}^{- 1}$

10x

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5000x

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200x

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Solution

The correct option is B 5000x
As we know
$M o l a r i t y = \frac{M o l e s o f s o l u t e}{L i t e r s o f s o l u t i o n}$
$M = \frac{0.1}{\frac{500}{1000}}$
$M = \frac{0.1 \times 1000}{500}$
$M = 0.2$
Now,
$Λ_{m} = \frac{K \times 1000}{M}$
where
K=specific conductance
M=molar concentration
$Λ_{m} = \frac{x \times 1000}{0.2}$
$Λ_{m} = 5000 x$

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