JEE CHEMISTRY MCQ: Equilibrium

1 Hard

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Analyzing +4 | -1

A volume of

1.0 \mathrm{~L}

of solution which was in equilibrium with solid mixture of

{AgCl}

and

{Ag}_2 {CrO}_4

was found to contain

1 \times 10^{-4}

moles of

{Ag}^{+}

ions,

1.0 \times 10^{-6}

moles of

{Cl}^{-}

ions and

8.0 \times 10^{-4}

moles of

{CrO}_4{ }^{2-}

ions. The

{Ag}^{+}

ions are added slowly to the above mixture (keeping the volume constant) till

8.0 \times 10^{-7}

moles of

{AgCl}

got precipitated. How many moles of

{Ag}_2 {CrO}_4

were precipitated simultaneously?

View answer and explanation

Correct option: A

7.68 \times 10^{-4}

Option A (

7.68 \times 10^{-4}

) is correct because it calculates the amount of

Ag_2CrO_4

precipitated based on the change in chloride concentration and the solubility product constants.

More Questions from Equilibrium

Practice All Chapter MCQs →

\mathrm{K}_1

\mathrm{K}_2

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

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

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

\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}

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

\mathrm{pH}

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

\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

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

\text{O}_2