1.

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

Answer»

0.1 ML
0.4 mL
2.08mL
49.6 g

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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