Normality And Equivalent Concept
A 250 mL solution contains 4.9 g of sulfuric acid ((\text{H}_2\text{SO}_4)). What is the normality of this solution? (Molar mass of (\text{H}_2\text{SO}_4 = 98) g/mol; (n)-factor for (\text{H}_2\text{SO}_4) in acid-base reaction = 2)
Select the correct option:
Solution
0.4 N
Normality is defined as the number of gram equivalents of solute dissolved per litre of solution. The gram equivalent mass (equivalent weight) of an acid is its molar mass divided by the n-factor (basicity or number of replaceable protons). For (\text{H}_2\text{SO}4), equivalent weight (= 98/2 = 49) g/eq. Gram equivalents in 4.9 g: (n{\text{eq}} = 4.9/49 = 0.1) equivalents. Volume (= 250 , \text{mL} = 0.250) L. Normality (N = 0.1/0.250 = 0.4) N. Alternatively, normality (= \text{molarity} \times n\text{-factor}). Molarity (= (4.9/98)/0.250 = 0.05/0.250 = 0.2) M. Normality (= 0.2 \times 2 = 0.4) N. Option 0.2 N is the molarity, not the normality — the n-factor of 2 has not been applied. Option 0.5 N would require 6.125 g of (\text{H}_2\text{SO}_4) in 250 mL, not 4.9 g. Option 1.0 N would require 12.25 g in 250 mL. Normality is particularly important in acid-base titrimetry and is a higher-order concept tested in JEE Advanced. Plausibility check: since (N = M \times n)-factor (= 0.2 \times 2 = 0.4) N, and normality must always be (\geq) molarity for polyprotic acids, 0.4 N (>) 0.2 M is consistent.
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About This Question
- Subject
- chemistry
- Chapter
- some basic concepts in chemistry
- Topic
- normality and equivalent concept
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
0.4 N
Normality is defined as the number of gram equivalents of solute dissolved per litre of solution. The gram equivalent mass (equivalent weight) of an acid is its molar mass divided by the n-factor (basicity or number of replaceable protons). For (\text{H}_2\text{SO}4), equivalent weight (= 98/2 = 49) g/eq. Gram equivalents in 4.9 g: (n{\text{eq}} = 4.9/49 = 0.1) equivalents. Volume (= 250 , \text{mL} = 0.250) L. Normality (N = 0.1/0.250 = 0.4) N. Alternatively, normality (= \text{molarity} \times n\text{-factor}). Molarity (= (4.9/98)/0.250 = 0.05/0.250 = 0.2) M. Normality (= 0.2 \times 2 = 0.4) N. Option 0.2 N is the molarity, not the normality — the n-factor of 2 has not been applied. Option 0.5 N would require 6.125 g of (\text{H}_2\text{SO}_4) in 250 mL, not 4.9 g. Option 1.0 N would require 12.25 g in 250 mL. Normality is particularly important in acid-base titrimetry and is a higher-order concept tested in JEE Advanced. Plausibility check: since (N = M \times n)-factor (= 0.2 \times 2 = 0.4) N, and normality must always be (\geq) molarity for polyprotic acids, 0.4 N (>) 0.2 M is consistent.
This hard difficulty chemistry question is from the chapter some basic concepts in chemistry, covering the topic of normality and equivalent concept. It appeared in the 2025 exam.
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