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What will be the value of $\land_{c}$ for 0.001M aqueous solution of NH₃ ? For $N H_{3} K_{b} = 1.6 \times 10^{- 5}$ and $\land_{0} = 2.38 \times 10^{- 2} S m^{2} m o l^{- 1}$

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$3.01 \times 10^{- 3} S m^{2} m o l e^{- 1}$

$0.1265 S m^{2} m o l e^{- 1}$

$301 \times 10^{- 3} S m^{2} m o l e^{- 1}$

$2.38 \times 10^{- 2} S m^{2} m o l e^{- 1}$

answer is A.

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Detailed Solution

$α = \sqrt{\frac{K_{b}}{C}}; α = \frac{\land_{c}}{\land_{0}}$

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