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Some Basic Concepts in Chemistry: Chemistry | JEE Main The strength of 50 volume solution of hydrogen peroxide is
___________ g/L \text{\_\_\_\_\_\_\_\_\_\_\_}\text{ g/L} ___________ g/L (Nearest integer).
Given:
Molar mass of
H 2 O 2 \mathrm{H}_2\mathrm{O}_2 H 2 O 2 is
34 g mol − 1 34\text{ g mol}^{-1} 34 g mol − 1
Molar volume of gas at
S T P = 22.7 L \mathrm{STP} = 22.7\text{ L} STP = 22.7 L .
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Choose one answer I’d rather see the answer directly Step-by-step solution View Correct answer
The strength of a 50 volume H 2 O 2 \mathrm{H}_2\mathrm{O}_2 H 2 O 2 22.7 L 22.7\text{ L} 22.7 L 150 g/L 150\text{ g/L} 150 g/L Option analysis
Why each option works or fails
Step Working
01 given Volume strength of H 2 O 2 = 50 volume \mathrm{H_2O_2} = 50\text{ volume} H 2 O 2 = 50 volume H 2 O 2 = 34 g mol − 1 \mathrm{H_2O_2} = 34\text{ g mol}^{-1} H 2 O 2 = 34 g mol − 1 S T P = 22.7 L \mathrm{STP} = 22.7\text{ L} STP = 22.7 L
02 strategise From the decomposition reaction: 2 H 2 O 2 ( a q ) → 2 H 2 O ( l ) + O 2 ( g ) 2\mathrm{H_2O_2(aq)} \rightarrow 2\mathrm{H_2O(l)} + \mathrm{O_2(g)} 2 H 2 O 2 ( aq ) → 2 H 2 O ( l ) + O 2 ( g ) 2 moles 2\text{ moles} 2 moles H 2 O 2 \mathrm{H_2O_2} H 2 O 2 1 mole 1\text{ mole} 1 mole O 2 \mathrm{O_2} O 2 22.7 L 22.7\text{ L} 22.7 L 1 L 1\text{ L} 1 L 1 M H 2 O 2 1\text{ M } \mathrm{H_2O_2} 1 M H 2 O 2 11.35 L 11.35\text{ L} 11.35 L O 2 \mathrm{O_2} O 2 Volume strength = 11.35 × Molarity \text{Volume strength} = 11.35 \times \text{Molarity} Volume strength = 11.35 × Molarity Strength in g/L = Molarity × Molar mass \text{Strength in g/L} = \text{Molarity} \times \text{Molar mass} Strength in g/L = Molarity × Molar mass
03 execute Calculate Molarity: M = 50 11.35 M = \frac{50}{11.35} M = 11.35 50 g/L = M × 34 = 50 11.35 × 34 = 1700 11.35 ≈ 149.78 g/L ≈ 150 g/L \text{g/L} = M \times 34 = \frac{50}{11.35} \times 34 = \frac{1700}{11.35} \approx 149.78\text{ g/L} \approx 150\text{ g/L} g/L = M × 34 = 11.35 50 × 34 = 11.35 1700 ≈ 149.78 g/L ≈ 150 g/L
✓ verify Checking rounding to nearest integer: 1700 / 11.35 = 149.7797... 1700 / 11.35 = 149.7797... 1700/11.35 = 149.7797...
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Question type Numerical
Exam relevance JEE Main · Chemistry
Concepts assessed Chemistry
Academic status Reviewed by official_key
Source pyq
Editorial review 9 September 2026 Quick checks
Students also ask Why do we divide 22.7 by 2 to get 11.35? Because 2 moles 2\text{ moles} 2 moles H 2 O 2 \mathrm{H_2O_2} H 2 O 2 1 mole 1\text{ mole} 1 mole 22.7 L 22.7\text{ L} 22.7 L O 2 \mathrm{O_2} O 2 1 mole 1\text{ mole} 1 mole H 2 O 2 \mathrm{H_2O_2} H 2 O 2 0.5 mole 0.5\text{ mole} 0.5 mole 11.35 L 11.35\text{ L} 11.35 L O 2 \mathrm{O_2} O 2
Answer The strength of a 50 volume H 2 O 2 \mathrm{H}_2\mathrm{O}_2 H 2 O 2 22.7 L 22.7\text{ L} 22.7 L 150 g/L 150\text{ g/L} 150 g/L
Why each option works or fails Step-by-step solution given: Volume strength of H 2 O 2 = 50 volume \mathrm{H_2O_2} = 50\text{ volume} H 2 O 2 = 50 volume H 2 O 2 = 34 g mol − 1 \mathrm{H_2O_2} = 34\text{ g mol}^{-1} H 2 O 2 = 34 g mol − 1 S T P = 22.7 L \mathrm{STP} = 22.7\text{ L} STP = 22.7 L strategise: From the decomposition reaction: 2 H 2 O 2 ( a q ) → 2 H 2 O ( l ) + O 2 ( g ) 2\mathrm{H_2O_2(aq)} \rightarrow 2\mathrm{H_2O(l)} + \mathrm{O_2(g)} 2 H 2 O 2 ( aq ) → 2 H 2 O ( l ) + O 2 ( g ) 2 moles 2\text{ moles} 2 moles H 2 O 2 \mathrm{H_2O_2} H 2 O 2 1 mole 1\text{ mole} 1 mole O 2 \mathrm{O_2} O 2 22.7 L 22.7\text{ L} 22.7 L 1 L 1\text{ L} 1 L 1 M H 2 O 2 1\text{ M } \mathrm{H_2O_2} 1 M H 2 O 2 11.35 L 11.35\text{ L} 11.35 L O 2 \mathrm{O_2} O 2 Volume strength = 11.35 × Molarity \text{Volume strength} = 11.35 \times \text{Molarity} Volume strength = 11.35 × Molarity Strength in g/L = Molarity × Molar mass \text{Strength in g/L} = \text{Molarity} \times \text{Molar mass} Strength in g/L = Molarity × Molar mass execute: Calculate Molarity: M = 50 11.35 M = \frac{50}{11.35} M = 11.35 50 g/L = M × 34 = 50 11.35 × 34 = 1700 11.35 ≈ 149.78 g/L ≈ 150 g/L \text{g/L} = M \times 34 = \frac{50}{11.35} \times 34 = \frac{1700}{11.35} \approx 149.78\text{ g/L} \approx 150\text{ g/L} g/L = M × 34 = 11.35 50 × 34 = 11.35 1700 ≈ 149.78 g/L ≈ 150 g/L verify: Checking rounding to nearest integer: 1700 / 11.35 = 149.7797... 1700 / 11.35 = 149.7797... 1700/11.35 = 149.7797... Shortcut: When to use it: When formula Volume strength = 11.35 × M is not memorised and needs derivation from definition.
given: 1 L 1\text{ L} 1 L H 2 O 2 \mathrm{H_2O_2} H 2 O 2 50 L 50\text{ L} 50 L O 2 \mathrm{O_2} O 2 = 22.7 L/mol = 22.7\text{ L/mol} = 22.7 L/mol H 2 O 2 = 34 g/mol \mathrm{H_2O_2} = 34\text{ g/mol} H 2 O 2 = 34 g/mol
strategise: Find moles of O 2 \mathrm{O_2} O 2 n O 2 = 50 22.7 n_{\mathrm{O_2}} = \frac{50}{22.7} n O 2 = 22.7 50 H 2 O 2 = 2 × n O 2 \mathrm{H_2O_2} = 2 \times n_{\mathrm{O_2}} H 2 O 2 = 2 × n O 2 H 2 O 2 = n H 2 O 2 × 34 \mathrm{H_2O_2} = n_{\mathrm{H_2O_2}} \times 34 H 2 O 2 = n H 2 O 2 × 34
execute: Mass of H 2 O 2 = 2 × 50 22.7 × 34 = 3400 22.7 ≈ 149.78 g \text{Mass of } \mathrm{H_2O_2} = 2 \times \frac{50}{22.7} \times 34 = \frac{3400}{22.7} \approx 149.78\text{ g} Mass of H 2 O 2 = 2 × 22.7 50 × 34 = 22.7 3400 ≈ 149.78 g
verify: 3400 22.7 = 149.7797... \frac{3400}{22.7} = 149.7797... 22.7 3400 = 149.7797...