What volume of O2(g) measured at 1 atm and 273 K will be formed by action of 100 mL of 0.5N KMnO4 on hydrogen peroxide in an acid solution ? The skeleton equation for the reaction isKMnO4+H2SO4+H2O2⟶K2SO4+MnSO4+O2+H2O

2KMnO4​+3H2​SO4​+5H2​O2​⟶K2​SO4​+2MnSO4​+5O2​+8H2​O⟶(1)Normality of KMnO4​ solution=0.5⇒ Molarity of KMnO4​ solution=50/.5​=0.1M[∵ Normality=Molarity × n factorHere, for KMnO4​ n factor is 5 [as e− change is 5]Now, Molarity =0.1M, which means0.1 moles of KMnO4​ dissolved in 1000 ml solution∴ 1n 100ml solution, moles of KMnO4​ is 100×0.1/ 1000  ​=0.01 molesFrom (1) , it is clear that2 moles KMnO4​ produce 5 moles O2​⇒0.01 moles KMnO4​ will give 0.025 moles of O2​∴ Molar volume of 0.025 moles of O2​ gas is as 1 mole=22.4L of O2​⇒0.025 moles=0.025×22.4L=0.56 L