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Question
10 mL of a solution of H2O2 of 10 volume strength decolourises 100 mL of KMnO4 solution acidified with dil. H2SO4. The amount of KMnO4 in the given solution is (K=39,Mn=55) :
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Solution
The correct option is B 0.564 g
Volume of O2 at STP =10 mL ×10 V=100 mL
22400 mL of O2 at STP =1 mol =4 eq.
100 mL of O2 at STP =422400×100=156 eq.
Eq. of KMnO4= Eq. of O2
=156 Eq. of KMnO4 =156×31.5 g of KMnO4 =0.564 g
(Equivalent weight of KMnO4 in acidic medium =Mw5=31.5)
Volume of O2 at STP =10 mL ×10 V=100 mL
22400 mL of O2 at STP =1 mol =4 eq.
100 mL of O2 at STP =422400×100=156 eq.
Eq. of KMnO4= Eq. of O2
=156 Eq. of KMnO4 =156×31.5 g of KMnO4 =0.564 g
(Equivalent weight of KMnO4 in acidic medium =Mw5=31.5)
100 mL of O2 at STP =422400×100=156 eq.
Eq. of KMnO4= Eq. of O2
=156 Eq. of KMnO4 =156×31.5 g of KMnO4 =0.564 g
(Equivalent weight of KMnO4 in acidic medium =Mw5=31.5)
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