A) \[\left[ Cr{{\left( {{H}_{2}}O \right)}_{6}} \right]C{{l}_{3}}\]
B) \[\left[ CrCl{{\left( {{H}_{2}}O \right)}_{5}} \right]C{{l}_{2}}.{{H}_{2}}O\]
C) \[\left[ CrC{{l}_{2}}{{\left( {{H}_{2}}O \right)}_{4}} \right]Cl.2{{H}_{2}}O\]
D) \[\left[ Cr{{\left( {{H}_{2}}O \right)}_{3}}C{{l}_{3}} \right]\]
Correct Answer: B
Solution :
[b] \[\underset{0.1\,mol}{\mathop{CrC{{l}_{3}}.6{{H}_{2}}O}}\,+AgN{{O}_{3}}\left( excess \right)\xrightarrow{{}}\underset{28.7g}{\mathop{AgCl}}\,\] No. of moles of \[AgCl=\frac{28.7}{143.5}\approx 0.2\text{ }mol\]
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