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What volume of 0.1M KMnO_4is needed to oxidise 5mg of ferrous oxalate in acidc medium (MW of ferrous oxalate is 144) approximately 0.20 mL |
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Answer» Solution : Milli equivalents of `KMnO_(4) = V xx 0.1 xx 5` milli equivalents of `FeC_(2)O_(4) = ("Wt in g")/("Ew") xx 1000` n-factor for `FeC_(2)O_(4)=3` milli equivalents of `KMnO_(4) = 5/(144//3) = (5xx3)/(144)` `:.` milliequivalents of `KMnO_4` = milli equivalents of `FeC_(2)O_4` `:. V xx 0.1 xx 5 =(5xx3)/144 :. V = 0.20 mL` |
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