Home
Class 12
CHEMISTRY
10 g of oxalate was dissolved in 300 mL ...

10 g of oxalate was dissolved in 300 mL of solution. This solution required 250 mL of `(M/10)KMnO_(4)`in acidic medium for complete oxidation. The percentage purity of oxalate ion in the salt is :

A

0.55

B

0.578

C

0.45

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
A
Number of equivalents of oxalate = Number of equivalents of `KMnO_(4)=(250)/(1000)xx(5)/(10)=(1)/(8)`
implies `(1)/(8)=(g)/(88)/(2)implies g=5.5gram` therefore %purity=`(5.5)/(10)xx100=55%`
Promotional Banner

Similar Questions

Explore conceptually related problems

0.3 g of an oxalate salts was dissolved in 100 mL solution. The solution required 90 mL of N//20 KMnO_(4) for complete oxidation. The % of oxalate ion in salt is:

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 .

If 1.26 grams of oxalic acid is dissolved in 250 ml of solution then its normality is

The weight of oxalic acid required to reduce 80 ml of 0.4 M KMnO_(4) in acidic medium is