100 mL of 0.1 M NaHC 2 O 4 is neutralised by V1 mL of 0.1 M NaOH and V2 mL of a M KMnO 4 separately, then for complete neutralization :

Question

100 mL of 0.1 M NaHC 2 O 4 is neutralised by V1 mL of 0.1 M NaOH and V2 mL of a M KMnO 4 separately, then for complete neutralization : (A) Volume of NaOH required =200 mL (B) If M of KMnO 4 is 0.1 M then (V1/V2)=5: 2 (C) If M of KMnO 4 is 0.1 M then V2=40 mL (D) If M of KMnO 4 is 0.2 M then V2=2 mL

Tardigrade Admin · Accepted Answer

Correct answer is (b) If M of KMnO 4 is 0.1 M then (V1/V2)=5: 2Correct answer is (c) If M of KMnO 4 is 0.1 M then V2=40 mL

Q. $100 m L$ of $0.1 M N a H C_{2} O_{4}$ is neutralised by $V_{1} m L$ of $0.1 M N a O H$ and $V_{2} m L$ of a $M K M n O_{4}$ separately, then for complete neutralization :

Volume of $N a O H$ required $= 200 m L$

If $M$ of $K M n O_{4}$ is $0.1 M$ then $\frac{V _{1}}{V _{2}} = 5 : 2$

If $M$ of $K M n O_{4}$ is $0.1 M$ then $V_{2} = 40 m L$

If $M$ of $K M n O_{4}$ is $0.2 M$ then $V_{2} = 2 m L$

Correct answer is (b) If $M$ of $K M n O_{4}$ is $0.1 M$ then $\frac{V _{1}}{V _{2}} = 5 : 2$ Correct answer is (c) If $M$ of $K M n O_{4}$ is $0.1 M$ then $V_{2} = 40 m L$

Q. 100mL of 0.1MNaHC2​O4​ is neutralised by V1​mL of 0.1MNaOH and V2​mL of a MKMnO4​ separately, then for complete neutralization :

Volume of NaOH required =200mL

If M of KMnO4​ is 0.1M then V2​V1​​=5:2

If M of KMnO4​ is 0.1M then V2​=40mL

If M of KMnO4​ is 0.2M then V2​=2mL