Molar conductance of 0.2 M weak mono protic acid HA is 4.95 mho cm2 mole−1. Molar conductance of 0.002 M HA at the same temperature would be nearly ( 'α' is too small compared to 1)

α1α2=∧1∧2....(1)α=KaC......(2)α1α2=C2C1.....(3)Compairing 1 and 3 ∧1∧2=C2C14.95∧2=0.0020.24.95∧2=110∧2=49.5mhocm2mole−1

Molar conductance of 0.2 M weak mono protic acid HA is 4.95 mho cm2 mole−1. Molar conductance of 0.002 M HA at the same temperature would be nearly ( 'α' is too small compared to 1)

α1α2=∧1∧2....(1)α=KaC......(2)α1α2=C2C1.....(3)Compairing 1 and 3 ∧1∧2=C2C14.95∧2=0.0020.24.95∧2=110∧2=49.5mhocm2mole−1

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Molar conductance of 0.2 M weak mono protic acid HA is 4.95 mho cm² mole⁻¹. Molar conductance of 0.002 M HA at the same temperature would be nearly ( ' $α$ ' is too small compared to 1)

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495 mho cm² mole^-1

4.95 mho cm² mole^-1

49.5 mho cm² mole^-1

0.0495 moh cm² mole^-1

answer is C.

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Detailed Solution

$\begin{array}{l} \frac{α_{1}}{α_{2}} = \frac{\land_{1}}{\land_{2}} . . . . (1) \\ α = \sqrt{\frac{K_{a}}{C}} . . . . . . (2) \\ \frac{α_{1}}{α_{2}} = \sqrt{\frac{C_{2}}{C_{1}}} . . . . . (3) \end{array}$
Compairing 1 and 3
$\begin{array}{l} \frac{\land_{1}}{\land_{2}} = \sqrt{\frac{C_{2}}{C_{1}}} \\ \frac{4.95}{\land_{2}} = \sqrt{\frac{0.002}{0.2}} \\ \frac{4.95}{\land_{2}} = \frac{1}{10} \\ \land_{2} = 49.5 mho {cm}^{2} {mole}^{- 1} \end{array}$