Question

The volume of 0.1 M oxalic acid that can be completely oxidised by 20 mL of 0.025 M $KMn{O}_4$ solution is:

• A. $25mL$
• B. $12.5mL$
• C. $37.5mL$
• D. $125mL$

Answer 1

Answer: (B)

$12.5mL$

Given $0.025MK{M}_n{O}_4$.

$\therefore N\left(KMn{O}_4\right)=\frac{158}{31.6}\times 0.025=0.125N$

Eq. wt of oxalic acid $=\frac{90}{2}=45$

$N\left(oxalicacid\right)=\frac{90}{45}\times 0.1=0.2N$

$\therefore (NV{)}_{\text{(oxalic acid)}}=(NV{)}_{KMn{O}_4}$

$V_{\text{oxalic acid}}=\frac{(NV{)}_{KMn{O}_4}}{N_{oxalicacid}}$

$=\frac{0.125\times 20}{0.2}$

$V_{\text{oxalic acid}}=12.5mL$

$\text{so the volume of oxalic acid is 12.5mL}$

Hence, the correct option is $\text{B}$