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 \[C{{l}^{-}}\] ion. \[\underset{\begin{smallmatrix} Initial \\ mmol \\ After\,ionisation \end{smallmatrix}}{\mathop{[Cr{{({{H}_{2}}O)}_{4}}}}\,\underset{\begin{smallmatrix} 100\times 0.01 \\ =1mmol \\ 0 \end{smallmatrix}}{\mathop{C{{l}_{2}}]Cl}}\,\xrightarrow[{}]{{}}\underset{\begin{smallmatrix} 0 \\ \\ 1mmol \end{smallmatrix}}{\mathop{[Cr{{({{H}_{2}}O)}_{4}}}}\,C{{l}_{2}}]\underset{\begin{smallmatrix} 0 \\ \\ 1mmol \end{smallmatrix}}{\mathop{^{+}+C{{l}^{-}}}}\,\]One mole of \[C{{l}^{-}}\] ions react with only 1 mole of \[AgN{{O}_{3}}\] molecule to produce 1 mole of \[AgCl\]. \[\therefore \]1 mmol or \[1\times {{10}^{-3}}\] mole reacts with \[AgN{{O}_{3}}\] to give \[AgCl\] \[=\frac{1\times 1\times {{10}^{-3}}}{1}={{10}^{-3}}or\,0.001\,mol\,AgCl\]
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