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20mL of `H_(2)O_(2)` after acidification with dilute `H_(2)SO_(4)` required 30mL of N/12 `KMnO_(4)` for complete oxidation. Calculate the percentage of `H_(2)O_(2)` in the solution. Equivalent mass of `H_(2)O_(2)=17`. |
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Answer» Correct Answer - 0.2125 `30 mL (N)/(12) KMnO_(4) -= 20 mL N_(x) H_(2)O_(2)` Normality of `H_(2)O_(2)`.soln., `Nx = (30)/(12 xx 20) = 1/8` Strength = Normality `xx` Eq. mass `= 1/8 xx 17 = 2.125 gL^(-1)` `% = (2.15)/(1000) xx 100 = 0.2125` |
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