0.3 g of an oxalate salt 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

Question

0.3 g of an oxalate salt 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 (A) 33 % (B) 66 % (C) 70 % (D) 40 %

Tardigrade Admin · Accepted Answer

Meq. of KMnO 4= M eq. of C 2 O 4-2 90 × (1/20)=100 × N C 2 O 4-2 =(9/2 × 2)=(9/4) Wt of oxalate =(9/4) × 88 × 10-3=22 × 9 × 10-3=198 × 10-3 % C 2 O 4-2=(.198/.300) × 100=66 %

Q. $0.3 g$ of an oxalate salt was dissolved in $100 m L$ solution. The solution required $90 m L$ of $N /20 K M n O_{4}$ for complete oxidation. The $%$ of oxalate ion in salt is

$33%$

$66%$

$70%$

$40%$

Q. 0.3g of an oxalate salt was dissolved in 100mL solution. The solution required 90mL of N/20KMnO4​ for complete oxidation. The % of oxalate ion in salt is

33%

66%

70%

40%

Meq. of KMnO4​=M eq. of C2​O4−2​ 90×201​=100×NC2​O4−2​​ =2×29​=49​ Wt of oxalate =49​×88×10−3=22×9×10−3=198×10−3 %C2​O4−2​=.300.198​×100=66%

Q. $0.3 g$ of an oxalate salt was dissolved in $100 m L$ solution. The solution required $90 m L$ of $N /20 K M n O_{4}$ for complete oxidation. The $%$ of oxalate ion in salt is

$33%$

$66%$

$70%$

$40%$