JEE CHEMISTRY MCQ: Equilibrium

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Application +4 | -1

For the equilibrium

{SrCl}_2 \cdot 6 {H}_2{O}(\mathrm{s}) \rightleftharpoons {SrCl}_2 \cdot 2 {H}_2 {O}(\mathrm{s})+4 {H}_2 {O}(\mathrm{g}),

K_{{P}}=8.1 \times 10^{-7} \mathrm{~atm}^4

27^{\circ} \mathrm{C}.

1.642 \mathrm{~L}

of air saturated with water vapour at

27^{\circ} \mathrm{C}

is exposed to a large quantity of

{SrCl}_2 \cdot 2 {H}_2 {O} \text{(s)},

then what mass of water vapour will be absorbed? Saturated vapour pressure of water at

27^{\circ} \mathrm{C}=30.4 \text{ torr}.

View answer and explanation

Correct option: A

12 \mathrm{~mg}

Option A (12 mg) is correct because it calculates the mass of water vapor absorbed by SrCl2.2H2O(s) by considering the difference between the actual and equilibrium partial pressures of water vapor.

More Questions from Equilibrium

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The indicator used in titrating oxalic acid with caustic soda solution is:

\mathrm{K}_1

and

\mathrm{K}_2

are equilibrium constant for reactions (i) and (ii)

\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\quad\ldots

(i)

\mathrm{NO}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{~N}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})\quad\ldots

(ii)

Then,

K{a_1},\,\,\,K{a_2}\,\,\,\mathrm{and}\,\,\,K{a_3}

are the respective ionisation constants for the following reactions

\begin{array}{l} {H_2}S\leftrightharpoons{H^ + } + H{S^ - }\\ H{S^ - }\leftrightharpoons{H^ + } + {S^{2 - }}\\ {H_2}S\leftrightharpoons2{H^ + } + {S^{2 - }} \end{array}

The correct relationship between

K{a_1},\,\,K{a_2}\,,K{a_3}

is The

\mathrm{pH}

0.1 \mathrm{~M}-\mathrm{N}_2 \mathrm{H}_4

solution is (For

\mathrm{N}_2 \mathrm{H}_4,

K_{{b}_1}=3.6 \times 10^{-6}, K_{{b}_2}=6.4 \times 10^{-12}, \log 2=0.3

\log 3=0.48

) Reaction

{\text{Ba}}{{\text{O}}_2}(s) \rightleftharpoons {\text{BaO}}(s) + {{\text{O}}_2}(g);\Delta H = - {\text{ve}}.

In equilibrium condition pressure of

\text{O}_2

depends on: A : At equilibrium concentration of the reactant and product does not change with time for a chemical reaction.
R: The rate of reaction is zero at equilibrium.