Autoionisation of liquid NH3 occurs as2NH3⇌NH4++NH2− with KNH3=NH4+NH2−=10−30 at −50∘CNumber of amide ions, NH2−, present per mm3 of pure liquid NH3 is

NH2−=NH4+=KNH3=10−15M=10−15molL−1 10−15molL−1=10−15mol1000cm3=10−15mol106mm3 =10−15×6.02×1023106NH2−ion /mm3=602

Autoionisation of liquid NH3 occurs as2NH3⇌NH4++NH2− with KNH3=NH4+NH2−=10−30 at −50∘CNumber of amide ions, NH2−, present per mm3 of pure liquid NH3 is

NH2−=NH4+=KNH3=10−15M=10−15molL−1 10−15molL−1=10−15mol1000cm3=10−15mol106mm3 =10−15×6.02×1023106NH2−ion /mm3=602

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Autoionisation of liquid NH₃ occurs as
$2 {NH}_{3} ⇌ {NH}_{4}^{+} + {NH}_{2}^{-} with K_{{NH}_{3}} = [{NH}_{4}^{+}] [{NH}_{2}^{-}] = 10^{- 30} at - 50^{\circ} C$
Number of amide ions, ${NH}_{2}^{-},$ present per mm³ of pure liquid NH₃ is

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602

300

200

100

answer is A.

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Detailed Solution

$[{NH}_{2}^{-}] = [{NH}_{4}^{+}] = \sqrt{K_{{NH}_{3}}} = 10^{- 15} M = 10^{- 15} {molL}^{- 1} 10^{- 15} {molL}^{- 1} = \frac{10^{- 15} mol}{1000 {cm}^{3}} = \frac{10^{- 15} mol}{10^{6} {mm}^{3}} = \frac{10^{- 15} \times 6 .02 \times 10^{23}}{10^{6}} {NH}_{2}^{-} ion / {mm}^{3} = 602$