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Q.

0.3g of an oxalate salt was dissolved in 100 mL  solution. The solution required 90 mL of N/20  KMnO4 for complete oxidation. The percentage of oxalate ion in salt is:

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a

66%

b

70%

c

40%

d

33%

answer is B.

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Detailed Solution

The percentage of oxalate ion in salt is 66%.

Number of mili equivalent of  KMnO4=90×1/20=4.5 m equivalents

Number of mili equivalents of oxalate ions = number of mili equivalents of KMnO4=4.5 m equivalents
Number of equivalents  = Mass  equivalent mass  

& equivalent mass of oxalate ion =88/2=44

 Mass of oxalate =44×4.5×10-3grams

=198×10-3 g 
=0.198 g

 percentage in salt oxalate =0.1980.3×100=66%

Hence, option B is correct.

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