A) 23.4
B) 33.6
C) 43.5
D) 53.6
Correct Answer: C
Solution :
\[\underset{1\,\text{mole}\,\text{(M)}}{\mathop{Mn{{O}_{2}}(g)}}\,+4HCl(aq)\xrightarrow{\Delta }\] \[MnC{{l}_{2}}+2{{H}_{2}}O+\underset{(71\,g)}{\mathop{C{{l}_{2}}(g)}}\,\] \[\because \]71 got \[C{{l}_{2}}\]is displaced by M g of \[\text{Mn}{{\text{O}}_{\text{2}}}\] \[\therefore \]\[35.5\,g\]of \[C{{l}_{2}}\]is displaced by \[\frac{M\times 35.5}{71}\] \[=\frac{M}{2}g\]of \[Mn{{O}_{2}}\] \[\therefore \]Eq. wt. of \[Mn{{O}_{2}}=\frac{\text{Molecular}\,\text{weight}\,\text{of}\,\text{Mn}{{\text{O}}_{\text{2}}}}{\text{2}}\] \[=\frac{87}{2}=43.5\]You need to login to perform this action.
You will be redirected in
3 sec