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A solution of a 0.4 g sample of H_(2)O_(2) reactedwith 0.632g of KMnO_(4)in the pressure of sulphuricacid.Calculatedthe percentage purity of the sample of H_(2)O_(2) . |
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Answer» Solution :`{:(2KMnO_(4),+3H_(2)SO_(4),+5H_(2)O_(2) ,to K_(2)SO_(4),+2MnSO_(4),+8H_(2)O,+5O_(2)),("+7",,,,"+2",,):}` Eq . Wt . Of `KMnO_(4) = (" mol.wt")/("change in ON per mole ") = 158/5 = 31/6` Again from the above reaction we see that 2 moles of `KMnO_(4)` combine with 5 moles of `H_(2)O_(2)` Again from the above reaction we see that 2 moles of `KMnO_(4)` combine with 5 moles of `H_(2)O_(2)` or 1 equivalentof `KMnO_(4)` combines with `1/2 ` molesof `H_(2)O_(2)` ` :. ` equivalentwt. of `H_(2)O_(2) = (" mol .wt")/2 = 34/2 = 17 ` Now , m.e of `H_(2)O_(2)` = m.e of `KMnO_(4)` oreq. of `H_(2)O_(2)`= eq. of `KMnO_(4)` `x/17=(0.632)/(31.6) "" ` ....(EQN . 4I) x being the amount of `H_(2)O_(2)` is GRAMS , x = 0.34 g Percentage of `H_(2)O_(2)` in the sample = `(0.34)/(0.4) xx 100 = 85 % ` |
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