$0.5 g$ of an oxalate was dissolved in water and the solution made to $100 m L$ . On titration $10 m L$ of this solution required $15 m L$ of $\frac{N}{20} K M n O_{4}$ . Calculate the percentage of oxalate in the sample.

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Solution

$15 m L \frac{N}{20} K M n O_{4} = 10 \times$ Normality of oxalate solution.
Normality of oxalate solution $= \frac{15}{20} \times \frac{1}{20} = \frac{3}{40}$
Strength of oxalate solution $= N o r m a l i t y \times$ Eq. mass of oxalate
$= \frac{3}{40} \times 44 = 3.3 g / L$ [Eq. mass of $C_{2} O_{4}^{2 -} = \frac{88}{2} = 44]$
Amount of oxalate in $100 m L$ solution $= \frac{3.3}{1000} \times 100 = 0.33 g$
% of oxalate $= \frac{0.33}{0.5} \times 100 = 66.0$

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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 N20KMnO4. Calculate the percentage of oxalate in the sample.

15 mLN20KMnO4=10× Normality of oxalate solution.Normality of oxalate solution =1520×120=340Strength of oxalate solution =Normality× Eq. mass of oxalate=340×44=3.3 g/L [Eq. mass of C2O2−4=882=44]Amount of oxalate in 100 mL solution =3.31000×100=0.33 g% of oxalate =0.330.5×100=66.0

$0.5 g$ of an oxalate was dissolved in water and the solution made to $100 m L$ . On titration $10 m L$ of this solution required $15 m L$ of $\frac{N}{20} K M n O_{4}$ . Calculate the percentage of oxalate in the sample.