Question

5 gm of an oxalate was dissolved in water and the solution raised to 100 ml. 10 ml of this solution when titrated with n/20 kmno4 required 10 ml of kmno4. Calculate the % of oxalate ion in the sample.

Answer 1

Answer:

15 mL

20

N

KMnO

4

=10× Normality of oxalate solution.

Normality of oxalate solution =

20

15

×

20

1

=

40

3

Strength of oxalate solution =Normality× Eq. mass of oxalate

=

40

3

×44=3.3 g/L [Eq. mass of C

2

O

4

2−

=

2

88

=44]

Amount of oxalate in 100 mL solution =

1000

3.3

×100=0.33 g

% of oxalate =

0.5

0.33

×100=66.0

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