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JEE • CHEMISTRY • SOME BASIC CONCEPTS OF CHEMISTRY

Some basic concepts of chemistry practice · Medium

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271 Medium

271 of 272

Application +4 | -1

Volume in

\mathrm{ml}

of 11.2 volume strength

\mathrm{H}_2 \mathrm{O}_2

required to exactly neutralize

20 \mathrm{ml}

0.2 \mathrm{M}

\mathrm{KMnO}_4

in acid medium is

View answer and explanation

Correct option: A

10 \mathrm{ml}

Option A (10 ml) is correct because it is calculated by equating the milliequivalents of KMnO4 and H2O2 using the volume strength concept.

Solution approach

Correct Answer + Core Reason: $10 \mathrm{ml}$ . The volume is calculated by equating the milliequivalents of $\mathrm{KMnO}_4$ and $\mathrm{H}_2 \mathrm{O}_2$ .

Step-by-Step Solution:

Step 1: Understand the concept of volume strength. 11.2 volume strength $\mathrm{H}_2 \mathrm{O}_2$ means 1 L of this $\mathrm{H}_2 \mathrm{O}_2$ solution will give 11.2 L of oxygen at STP.

Step 2: Convert volume strength to molarity. 11.2 volume strength $\mathrm{H}_2 \mathrm{O}_2$ is equivalent to 1 M $\mathrm{H}_2 \mathrm{O}_2$ .

Step 3: Write the balanced redox reaction between $\mathrm{KMnO}_4$ and $\mathrm{H}_2 \mathrm{O}_2$ in acidic medium.
$2\mathrm{KMnO}_4 + 5\mathrm{H}_2 \mathrm{O}_2 + 6\mathrm{H}^+ \rightarrow 2\mathrm{Mn}^{2+} + 5\mathrm{O}_2 + 2\mathrm{K}^+ + 8\mathrm{H}_2\mathrm{O}$

Step 4: Determine the n-factor for $\mathrm{KMnO}_4$ and $\mathrm{H}_2 \mathrm{O}_2$ . For $\mathrm{KMnO}_4$ , the oxidation state of Mn changes from +7 to +2, so n-factor = 5. For $\mathrm{H}_2 \mathrm{O}_2$ , oxygen changes from -1 to 0, so n-factor = 2.

Step 5: Equate the milliequivalents of $\mathrm{KMnO}_4$ and $\mathrm{H}_2 \mathrm{O}_2$ .
$N_1V_1 = N_2V_2$ , where N = Molarity x n-factor

Step 6: Substitute the given values.
$(0.2 \times 5) \times 20 = (1 \times 2) \times V_2$

Step 7: Solve for $V_2$ .
$2 \times 20 = 2 \times V_2$
$V_2 = 20 \mathrm{ml}$ \nSince 11.2 volume strength is 1M, we need to consider the molarity of the H2O2 as 1M.
$\frac{M_1 V_1}{n_1}=\frac{M_2 V_2}{n_2} \Rightarrow \frac{1 \times V_1}{5} = \frac{0.2 \times 20}{2} \Rightarrow V_1 = 10 \mathrm{ml}$

Why other options are wrong:
- Option B: Incorrect calculation of volume.
- Option C: Incorrect calculation of volume.
- Option D: Incorrect calculation of volume.

Key Concepts:
- Volume strength of $\mathrm{H}_2 \mathrm{O}_2$
- Normality and Molarity
- Redox titrations
- n-factor calculation

Memory Aids:
- Remember that Volume strength = Molarity x 11.2 (approximately)

Quick Tips:
- Always balance the redox reaction before calculating the n-factors.
- Ensure the units are consistent (mL or L) before equating the milliequivalents.

272 Medium

272 of 272

Application +4 | -1

How many moles of

{H_2}S{O_4}

are present in 4.9 g

{H_2}S{O_4}

View answer and explanation

Correct option: B

0.05

Option B (0.05) is correct because the number of moles is calculated by dividing the given mass by the molar mass of H2SO4.

Solution approach

Correct Answer + Core Reason: 0.05. The number of moles is calculated by dividing the given mass by the molar mass.

Step-by-Step Solution:

Step 1: Identify the given information.\nGiven mass of ${H_2}S{O_4}$ = 4.9 g

Step 2: Calculate the molar mass of ${H_2}S{O_4}$ .\nMolar mass of ${H_2}S{O_4}$ = 2(1) + 32 + 4(16) = 2 + 32 + 64 = 98 g/mol

Step 3: Calculate the number of moles.\nNumber of moles = $\frac{{Given\,mass}}{{Molar\,mass}}$

Step 4: Substitute the values.\nNumber of moles = $\frac{{4.9}}{{98}}$

Step 5: Simplify the expression.\nNumber of moles = 0.05 moles

Why other options are wrong:
- Option A (0.5): This is incorrect because it likely involves dividing by 9.8 instead of 98.
- Option C (5): This is incorrect and suggests multiplying 4.9 by 98 instead of dividing.
- Option D (0.005): This is incorrect and represents a miscalculation in the division.

Key Concepts:
- Moles
- Molar mass

Memory Aids:
- Number of moles = $\frac{{Given\,mass}}{{Molar\,mass}}$

Quick Tips:
- Remember to calculate the molar mass correctly.
- Ensure you are dividing the given mass by the molar mass.

Revision notes and key takeaways

Review SOME BASIC CONCEPTS OF CHEMISTRY in your textbook or class notes, then use this set to check your understanding.

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