40 mL of x M KMnO 4 solution is required to react completely with 200 mL of 0.02 M oxalic acid solution in acidic medium. The value of x is

Question

40 mL of x M KMnO 4 solution is required to react completely with 200 mL of 0.02 M oxalic acid solution in acidic medium. The value of x is (A) 0.04 (B) 0.01 (C) 0.03 (D) 0.02

Tardigrade Admin · Accepted Answer

2 KMnO 4+3 H 2 SO 4+5 underset overset|COOHCOOH longrightarrow K 2 SO 4+2 MnSO 4+8 H 2 O +10 CO 2 ∵ (M1 V1/n1)( KMnO 4)=(M2 V2/n2) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/3eae4bcbbe33462405ed7a5227c2bd10-.png" /> ⇒ (x × 40 mL /2)=(0.02 × 200 mL /5) ∴ x=0.04 M

Q. $40 m L$ of $x M K M n O_{4}$ solution is required to react completely with $200 m L$ of $0.02 M$ oxalic acid solution in acidic medium. The value of $x$ is

0.04

0.01

0.03

0.02

$2 K M n O_{4} + 3 H_{2} S O_{4} + 5 C ∣ OO H COO H ⟶ K_{2} S O_{4} + 2 M n S O_{4} + 8 H_{2} O + 10 C O_{2}$

$∵ \frac{M _{1} V _{1}}{n _{1}} (K M n O_{4}) = \frac{M _{2} V _{2}}{n _{2}}$

$\Rightarrow \frac{x \times 40 m L}{2} = \frac{0.02 \times 200 m L}{5}$

$∴ x = 0.04 M$

Q. 40mL of xMKMnO4​ solution is required to react completely with 200mL of 0.02M oxalic acid solution in acidic medium. The value of x is