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Question

What volume of O2(g) measured at 1 atm and 273 K will be formed by the action of 100 ml of 0.5 N of KMnO4 on hydrogen peroxide in an acidic solution?
KMnO4+H2SO4+H2O2K2SO4+MnSO4+O2+H2O

A
0.12 L
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B
0.03 L
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C
0.56 L
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D
1.12 L
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Solution

The correct option is C 0.56 L
2KMnO4+3H2SO4+5H2O2K2SO4+2MnSO4+5O2+8H2O
Normality of KMnO4​ solution= 0.5 N
Since Normality = Mortality* n-factor
And n-factor for KMnO4 is 5
Molarity of KMnO4 solution= 50.5=0.1M
Now 0.1 M means 0.1 moles of KMnO4 are dissolved in 1000ml
so in 100 ml moles dissolved are 0.01 moles
Now according to the reaction, 2 moles of KMnO4 produce 5 moles of O2
hence 0.01 moles will produce 0.025 moles of O2
1 mole of oxygen gas occupies 22.4 L of gas
so 0.025 moles of oxygen gas will occupy
= 0.025*22.4 L = 0.56 L of O2

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