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Q.

An impure sample of sodium oxalate (Na2C2O4). Weighing 0.20 g is dissolved in aq. Solution of H2SO4  and solution is titrated at  700C required 45 ml of 0.02M KMnO4 solution. The end points is over run, and back titration is carried out with 10 ml of 0.1M oxalic acid solution. Find % purity of Na2C2O4 in the sample.
(round off to nearest integer)

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answer is 84.

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Detailed Solution

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Equivalents of  Na2C2O4= Equivalents of  KMnO4
= total equivalents of  KMnO4 excess equivalents of  KMnO4 reacted with  H2C2O4
1000×W×2134=45×0.02×510×0.1×2

 W=0.1675gm           % purity = 0.16750.2×100=83.75%84%

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