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5 mL solution of `H_2O_2` liberates 1.27 g of iodine from an acidified KI solution. What is the molarity of `H_2O_2`? |
Answer» `H_2O_2-=KI-=I_2` `mEq-=mEq-=mEq` `N_1xxV_1-=("weight of I_2")/(Ew(I_2))xx10^(3)mEq` `N_1xx5mL-=(1.27xx10^(3))/((127xx2)/(2))` (n-factor of `I_2=2`) `(2I^(ɵ)toI_2+2e^(-))` Normality of `H_2O_2-=2N` Molarity of `H_2O_2-=("normality")/(2)=(2)/(2)=1M` (n-factor of `H_2O_2=2)(2e^(-)+H_2O_2to2H_2O)` |
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