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Some Basic Concepts in Chemistry: Chemistry | JEE Main

Q: Why do we enter 224 instead of 2.24?

The stem specifies the answer in units of . Since the volume is , we rewrite it as , making the required integer 224.

The volume of hydrogen liberated at

\mathrm{STP}

by treating

2.4\mathrm{~g}

of magnesium with excess of hydrochloric acid is

\underline{\quad\quad} \times 10^{-2} \mathrm{~L}

. Given: Molar volume of gas is

22.4\mathrm{~L}

\mathrm{STP}

. Molar mass of magnesium is

24\mathrm{~g} \mathrm{~mol}^{-1}

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Step-by-step solutionView

Correct answer

The volume of hydrogen gas liberated at STP is $224 \times 10^{-2}\text{ L}$ , so the value to enter is $224$ .

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

Mass of $\mathrm{Mg} = 2.4\text{ g}$ , Molar mass of $\mathrm{Mg} = 24\text{ g mol}^{-1}$ , Molar volume of gas at $\mathrm{STP} = 22.4\text{ L mol}^{-1}$ . Excess $\mathrm{HCl}$ is used.

02find

Find the value of $N$ such that the volume of liberated $\mathrm{H}_2$ gas at $\mathrm{STP} = N \times 10^{-2}\text{ L}$ .

03visualise

Reaction: $\mathrm{Mg(s)} + 2\mathrm{HCl(aq)} \rightarrow \mathrm{MgCl}_2\mathrm{(aq)} + \mathrm{H}_2\mathrm{(g)}\uparrow$ . From stoichiometry, $1\text{ mol of } \mathrm{Mg}$ produces $1\text{ mol of } \mathrm{H}_2$ .

04strategise

1. Calculate moles of $\mathrm{Mg}$ : $n_{\mathrm{Mg}} = \frac{\text{mass}}{\text{molar mass}}$ . 2. Moles of $\mathrm{H}_2$ liberated: $n_{\mathrm{H}_2} = n_{\mathrm{Mg}}$ . 3. Volume of $\mathrm{H}_2$ at STP: $V = n_{\mathrm{H}_2} \times 22.4\text{ L}$ . 4. Express the answer in the form $N \times 10^{-2}\text{ L}$ .

05execute

Moles of $\mathrm{Mg} = \frac{2.4}{24} = 0.1\text{ mol}$ . Moles of $\mathrm{H}_2 = 0.1\text{ mol}$ . Volume of $\mathrm{H}_2 = 0.1 \times 22.4\text{ L} = 2.24\text{ L} = 224 \times 10^{-2}\text{ L}$ . Thus, the integer value to enter is $224$ .

✓verify

$224 \times 10^{-2}\text{ L} = 2.24\text{ L}$ . Since $2.4\text{ g}$ is exactly $0.1\text{ mol}$ of $\mathrm{Mg}$ , it generates $0.1\text{ mol}$ of $\mathrm{H}_2$ gas, which occupies $0.1 \times 22.4\text{ L} = 2.24\text{ L}$ . Result is consistent and dimensions match.

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Chemistry
Concepts assessed: Chemistry
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 7 September 2026

Quick checks

Students also ask

Why do we enter 224 instead of 2.24?

The stem specifies the answer in units of $\times 10^{-2}\text{ L}$ . Since the volume is $2.24\text{ L}$ , we rewrite it as $224 \times 10^{-2}\text{ L}$ , making the required integer 224.

Answer

The volume of hydrogen gas liberated at STP is $224 \times 10^{-2}\text{ L}$ , so the value to enter is $224$ .

Why each option works or fails

Step-by-step solution

given: Mass of $\mathrm{Mg} = 2.4\text{ g}$ , Molar mass of $\mathrm{Mg} = 24\text{ g mol}^{-1}$ , Molar volume of gas at $\mathrm{STP} = 22.4\text{ L mol}^{-1}$ . Excess $\mathrm{HCl}$ is used.
find: Find the value of $22.4\mathrm{~L}$ such that the volume of liberated $\mathrm{H}_2$ gas at $\mathrm{STP} = N \times 10^{-2}\text{ L}$ .
visualise: Reaction: $\mathrm{Mg(s)} + 2\mathrm{HCl(aq)} \rightarrow \mathrm{MgCl}_2\mathrm{(aq)} + \mathrm{H}_2\mathrm{(g)}\uparrow$ . From stoichiometry, $1\text{ mol of } \mathrm{Mg}$ produces $1\text{ mol of } \mathrm{H}_2$ .
strategise: 1. Calculate moles of $\mathrm{Mg}$ : $n_{\mathrm{Mg}} = \frac{\text{mass}}{\text{molar mass}}$ . 2. Moles of $\mathrm{H}_2$ liberated: $n_{\mathrm{H}_2} = n_{\mathrm{Mg}}$ . 3. Volume of $\mathrm{H}_2$ at STP: $V = n_{\mathrm{H}_2} \times 22.4\text{ L}$ . 4. Express the answer in the form $N \times 10^{-2}\text{ L}$ .
execute: Moles of $\mathrm{Mg} = \frac{2.4}{24} = 0.1\text{ mol}$ . Moles of $\mathrm{H}_2 = 0.1\text{ mol}$ . Volume of $\mathrm{H}_2 = 0.1 \times 22.4\text{ L} = 2.24\text{ L} = 224 \times 10^{-2}\text{ L}$ . Thus, the integer value to enter is $224$ .
verify: $224 \times 10^{-2}\text{ L} = 2.24\text{ L}$ . Since $2.4\text{ g}$ is exactly $0.1\text{ mol}$ of $\mathrm{Mg}$ , it generates $0.1\text{ mol}$ of $\mathrm{H}_2$ gas, which occupies $0.1 \times 22.4\text{ L} = 2.24\text{ L}$ . Result is consistent and dimensions match.

Some Basic Concepts in Chemistry: Chemistry | JEE Main

The volume of hydrogen gas liberated at STP is 224×10−2 L224 \times 10^{-2}\text{ L}224×10−2 L, so the value to enter is 224224224.

01given

02find

03visualise

04strategise

05execute

✓verify

Students also ask

The volume of hydrogen gas liberated at STP is $224 \times 10^{-2}\text{ L}$ , so the value to enter is $224$ .