0.3 g of an oxalate salt was dissolved in 100 mL solution. The solution required 90 mL of N/20 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 required 90 mL of N/20 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 KMnO4 = M eq. of C2O-24 90×(1/2)=100 × NC 2O4-2 Mmole of oxalate =(9/2×2)=(9/4) Weight of oxalate =(9/4) ×88×10-3=22 ×9×10-3 =198 ×10-3 % C2 O4-2 ×100=66 %

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

$33%$

$66%$

$70%$

$40%$

Q. 0.3 g of an oxalate salt was dissolved in 100 mL solution. The solution required 90 mL 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​ 90×21​=100×NC2​O4−2​​ Mmole of oxalate =2×29​=49​ Weight of oxalate =49​×88×10−3=22×9×10−3 =198×10−3 %C2​O4−2​×100=66%

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

$33%$

$66%$

$70%$

$40%$