SATHEE: Some Basic Concepts Of Chemistry Question 237

Some Basic Concepts Of Chemistry Question 237

Question: $K M n O_{4}$ reacts with oxalic acid according to the equation, $2 M n O_{4}^{-} + 5 C_{2} O_{4}^{2 -} + 16 H^{+} \to 2 M n^{2 +} + 10 C O_{2} + 8 H_{2} O$ , here 20 ml of 0.1 M $K M n O_{4}$ is equivalent to [CBSE PMT 1996]

Options:

A) 20 ml of 0.5 M $H_{2} C_{2} O_{4}$

B) 50 ml of 0.1 M $H_{2} C_{2} O_{4}$

C) 50 ml of 0.5 M $H_{2} C_{2} O_{4}$

D) 20 ml of 0.1 M $H_{2} C_{2} O_{4}$

Show Answer

Answer:

Correct Answer: B

Solution:

$K M n O_{4}$ Oxalic acid $\frac{M_{1} V_{1}}{n_{1}} =$ $\frac{M_{2} V_{2}}{n_{2}}$ ; $\frac{20 \times 0.1}{2} = \frac{M_{2} V_{2}}{5}$ ; $M_{2} V_{2} = 5$ .

Some Basic Concepts Of Chemistry Question 237

Question: $K M n O_{4}$ reacts with oxalic acid according to the equation, $2 M n O_{4}^{-} + 5 C_{2} O_{4}^{2 -} + 16 H^{+} \to 2 M n^{2 +} + 10 C O_{2} + 8 H_{2} O$ , here 20 ml of 0.1 M $K M n O_{4}$ is equivalent to [CBSE PMT 1996]

Options:

Answer:

Solution:

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Some Basic Concepts Of Chemistry Question 237

Question: KMnO4 reacts with oxalic acid according to the equation, 2MnO4−+5C2O42−+16H+→2Mn2++10CO2+8H2O , here 20 ml of 0.1 M KMnO4 is equivalent to [CBSE PMT 1996]

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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Question: $K M n O_{4}$ reacts with oxalic acid according to the equation, $2 M n O_{4}^{-} + 5 C_{2} O_{4}^{2 -} + 16 H^{+} \to 2 M n^{2 +} + 10 C O_{2} + 8 H_{2} O$ , here 20 ml of 0.1 M $K M n O_{4}$ is equivalent to [CBSE PMT 1996]