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Liquid $N H_{3}$ dissociates to a slight extent. At certain temperature, its self-dissociation constant $K_{S D C} (N H_{3}) is 10^{- 30}$ . The No. of $N H_{4}^{\oplus}$ ions present in pure liquid of 100 CC volume is $(N_{A} = 6 \times 10^{23})$

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$10^{- 15}$

$6 \times 10^{8}$

$6 \times 10^{7}$

$6 \times 10^{6}$

answer is C.

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Detailed Solution

$2 N H_{3} ⇌ N H_{2}^{-} + N H_{4}^{+} n - 2 x x x K_{S D C (N H_{3})} = [N H_{4}^{+}] [N H_{2}^{-}] = 10^{- 30} [N H_{4}^{+}] [N H_{2}^{-}] = 10^{- 30} [x] [x] = 10^{- 30} x = 10^{- 15} [N H_{4}^{+}] = 10^{- 15}$

Number of $N H_{4}^{\oplus}$ ions in 100 CC volume:

$10^{- 15} \times 6 \times 10^{23} \times \frac{100 c m^{3}}{1000 c m^{3} / L} = 6 \times 10^{7}$

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