Liquid ammonia ionises to a slight extent. At $ - 50^\circ C,$ its self ionisation constant, ${K_{{\rm{N}}{{\rm{H}}_3}}} = \left[ {{\rm{NH}}_4^ + } \right]\left[ {{\rm{NH}}_2^ - } \right] = {10^{ - 30}}.$ How many amide ions, are present per ${\rm{c}}{{\rm{m}}^3}$ of pure liquid ammonia? (Assume $N = 6.0 \times {10^{23}}$)

$6 \times {10^{ - 5}}\,{\rm{ions}}$

$6 \times {10^{ - 6}}\,{\rm{ions}}$

NEET UG CHEMISTRY MCQ: Equilibrium

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Application +4 | -1

Liquid ammonia ionises to a slight extent. At

- 50^\circ C,

  its self ionisation constant,

{K_{\mathrm{N}{\mathrm{H}_3}}} = \left[ {\mathrm{NH}_4^ + } \right]\left[ {\mathrm{NH}_2^ - } \right] = {10^{ - 30}}.

  How many amide ions, are present per

\mathrm{c}{\mathrm{m}^3}

of pure liquid ammonia? (Assume

N = 6.0 \times {10^{23}}

)

View answer and explanation

Correct option: B

6 \times {10^5}\,\mathrm{ions}

Option B (

6 \times 10^5

ions) is correct because it represents the calculated number of amide ions per cm³ based on the self-ionization constant.

Practice All Chapter MCQs →

Mg{(HC{O_3})_2}

MgC{O_3}

Mg{(OH)_2}

MgC{O_3}

K_{\mathrm{P}}

\mathrm{CaCO}_3

4.5 \times 10^{-2}

\mathrm{C}(\mathrm{s})+\mathrm{CO}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g}),

K_{\mathrm{P}}

2.0 \mathrm{~atm}.

\mathrm{CaO}

\mathrm{CaCO}_3

+

\leftrightharpoons

\mathrm{PC}{\mathrm{l}_5}\leftrightharpoons\mathrm{PC}{\mathrm{l}_3} + \mathrm{C}{\mathrm{l}_2}

{K_{\mathrm{eq}}} = 0.41

\mathrm{p}{K_a}