A) 0.001
B) 0.002
C) 0.003
D) 0.01
Correct Answer: A
Solution :
The formula of dichlorotetraqua chromium (III) chloride is \[[Cr{{({{H}_{2}}O)}_{4}}C{{l}_{2}}]Cl.\] On ionisation it generates only one CF ion. \[\underset{\text{After}\,\text{ionisation}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,}{\mathop{\underset{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,mmol\,\,\,\text{=}\,\text{1}\,\text{mmol}}{\mathop{\underset{\begin{smallmatrix} \,\text{Initial} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,100\,\,\times \,\,0.01\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \end{smallmatrix}}{\mathop{[Cr{{({{H}_{2}}O)}_{4}}C{{l}_{2}}]Cl}}\,}}\,}}\,\xrightarrow{{}}\underset{\text{1}\,\text{mmol}}{\mathop{\underset{0}{\mathop{{{[Cr{{({{H}_{2}}O)}_{4}}C{{l}_{2}}]}^{+}}}}\,}}\,+\underset{\text{1}\,\text{mmol}}{\mathop{\underset{0}{\mathop{C{{l}^{-}}}}\,}}\,\]One mole of \[C{{l}^{-}}\]ions react with only 1 mole of \[AgN{{O}_{3}}\]molecule to produce 1 mole of AgCI. \[\therefore \] 1 mmol or\[1\times {{10}^{-3}}\] mole reacts with \[AgN{{O}_{3}}\] to give AgCI
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