Question

0.5 g of an oxalate was dissolved in water and the solution made to 100 mL. On titration 10 mL of this solution required 15 mL of `(N)/(20)KMnO_(4)`. Calculate the percentage of oxalate in the sample .

Answer 1

`15 mL (N)/(20)KMnO_(4)=10xx`Normality of oxalate solution.
Normality of oxalate solution`= (15)/(10)xx(1)/(20)=(3)/(40)`
Strength of oxalate solution `= "Normality" xx " Eq. mass of oxalate"`
`=(3)/(40)xx44=3.3g//L [ " Eq. mass of " C_(2)O_(4)^(2-)=(88)/(2)=44]`
Amount of oxalate in 100 mL solution `= (3.3)/(1000)xx100=0.33 g`
% of oxalate `= (0.33)/(0.5)xx100=66.0`