A) 23.4
B) 33.6
C) 43.5
D) 53.6
Correct Answer: C
Solution :
\[\underset{1\text{ }mole\text{ (}M\text{)}}{\mathop{Mn{{O}_{2}}(g)}}\,+4HCl(aq)\xrightarrow{\Delta }\] \[MnC{{l}_{2}}+2{{H}_{2}}O+\underset{(71\,\,g)}{\mathop{C{{l}_{2}}(g)}}\,\] \[\because \] 71 g of \[C{{l}_{2}}\] is displaced by Mg of \[Mn{{O}_{2}}\] \[\therefore \] 35.5 g of \[C{{l}_{2}}\] is displaced by \[\frac{M\times 35.5}{71}\] \[=\frac{M}{2}g\] or \[Mn{{O}_{2}}\] \[\therefore \] Eq. wt. of \[Mn{{O}_{2}}=\frac{Molecular\text{ }weight\text{ }of\text{ }Mn{{O}_{2}}}{2}\] \[=\frac{87}{2}=43.5\]
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