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Question

For decolourisation of $$1.5$$ moles of $${ KMnO }_{ 4 }$$ in acidic medium the moles of $${ H }_{ 2 }{ O }_{ 2 }$$ requires$$:$$


A
32
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B
94
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C
154
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D
214
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Solution

The correct option is B $$\cfrac { 9 }{ 4 } $$
$$2KMn{ O }_{ 4 }+3{ H }_{ 2 }{ O }_{ 2 }\rightarrow 2Mn{ O }_{ 2 }+3{ O }_{ 2 }+2KOH+2{ H }_{ 2 }O$$
$$2$$ moles of $$KMn{ O }_{ 4 }\equiv 3$$ moles of $${ H }_{ 2 }{ O }_{ 2 }$$
$$\therefore \quad 1.5$$ moles of $$KMn{ O }_{ 4 }\equiv \cfrac { 3\times 1.5 }{ 2 } $$ moles of $${ H }_{ 2 }{ O }_{ 2 }$$
                                                $$\equiv \cfrac { 9 }{ 4 } $$ moles of $${ H }_{ 2 }{ O }_{ 2 }$$

Chemistry

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