0.3 g of an oxalate salt was dissolved in 100 ml solution. The solution requires 90 mL of (N/2)KMnO4 for complete oxidation. The % of oxalate ion in salt is

Question

0.3 g of an oxalate salt was dissolved in 100 ml solution. The solution requires 90 mL of (N/2)KMnO4 for complete oxidation. The % of oxalate ion in salt is (A) 33% (B) 66% (C) 70% (D) 40%

Tardigrade Admin · Accepted Answer

Meq. of oxalate = Meq. of KMnO4 (w/88/2)× 1000=90(1/20) ∴ w oxalate ion = 0.198 g ∴ of oxalate ion =(0.198/0.3)× 100=66%

Q. 0.3 g of an oxalate salt was dissolved in 100 ml solution. The solution requires 90 mL of $\frac{N}{2} K M n O_{4}$ for complete oxidation. The % of oxalate ion in salt is

33%

66%

70%

40%

Meq. of oxalate = Meq. of $K M n O_{4}$ $\frac{w}{88/2} \times 1000 = 90 \frac{1}{20}$ $∴$ w oxalate ion = 0.198 g $∴$ of oxalate ion $= \frac{0.198}{0.3} \times 100 = 66$

Q. 0.3 g of an oxalate salt was dissolved in 100 ml solution. The solution requires 90 mL of 2N​KMnO4​ for complete oxidation. The % of oxalate ion in salt is

33%

66%

70%

40%

Meq. of oxalate = Meq. of KMnO4​ 88/2w​×1000=90201​ ∴ w oxalate ion = 0.198 g ∴ of oxalate ion =0.30.198​×100=66

Q. 0.3 g of an oxalate salt was dissolved in 100 ml solution. The solution requires 90 mL of $\frac{N}{2} K M n O_{4}$ for complete oxidation. The % of oxalate ion in salt is

Meq. of oxalate = Meq. of $K M n O_{4}$ $\frac{w}{88/2} \times 1000 = 90 \frac{1}{20}$ $∴$ w oxalate ion = 0.198 g $∴$ of oxalate ion $= \frac{0.198}{0.3} \times 100 = 66$