The vapour density of completely dissociated ${\rm{N}}{{\rm{H}}_4}{\rm{Cl}}$ would be:

Slightly less than half of that of ammonium chloride

Double that of ammonium chloride

Determined by the amount of solid ammonium chloride used in the experiment

NEET UG CHEMISTRY MCQ: Equilibrium

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Understanding +4 | -1

The vapour density of completely dissociated

\mathrm{N}{\mathrm{H}_4}\mathrm{Cl}

would be:

View answer and explanation

Correct option: B

Half of that of ammonium chloride

Option B is correct because upon complete dissociation, one mole of NH4Cl yields two moles of products, halving the vapor density.

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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}