A) 2 g
B) 0.2 g
C) 0.17 g
D) 0.15 g
Correct Answer: B
Solution :
[b] \[mEq\,of\,Mn{{O}_{4}}^{-}\equiv mEq\,of\,{{H}_{2}}{{\text{O}}_{2}}\] \[10\times \frac{M}{5}\times 5\equiv mEq\,of\,{{H}_{2}}{{O}_{2}}\] \[\Rightarrow 10mEq\equiv mEq\,of\,{{H}_{2}}{{O}_{2}}\] Weight of \[{{H}_{2}}{{O}_{2}}=10\times {{10}^{-3}}\times \frac{34}{2}\times 0.17\] \[=0.2g\] Weight of impure \[{{H}_{2}}{{O}_{2}}\] \[=\frac{100}{85}\times 0.17=0.2g\]
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