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Some Basic Concepts of Chemistry question

2025 · 3 Apr · Shift 2 · Q4
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Some Basic Concepts of Chemistry question

2025 · 3 Apr · Shift 2 · Q4

JEE MainChemistrySome Basic Concepts of ChemistryMCQ+4 / −1
Mass of magnesium required to produce 220 mL of hydrogen gas at STP on reaction with excess of dil. HCl is Given: Molar mass of Mg is 24 g mol−124 \mathrm{~g} \mathrm{~mol}^{-1}24 g mol−1.
  1. A
    0.24 mg
  2. B
    235.7 g
  3. C
    2.444 g
  4. D
    236 mg
View written solutionFree

Correct answer: D

  1. Write the balanced reaction

Mg+2HCl→MgCl2+H2\mathrm{Mg + 2HCl \rightarrow MgCl_2 + H_2}Mg+2HCl→MgCl2​+H2​

From the equation,

  • 111 mole of Mg\mathrm{Mg}Mg produces 111 mole of H2\mathrm{H_2}H2​.
  1. Convert volume of hydrogen at STP to moles

At STP, 111 mole of any gas occupies 22.4 L22.4\,\text{L}22.4L.

Given volume of hydrogen:

220 mL=0.220 L220\,\text{mL} = 0.220\,\text{L}220mL=0.220L

So, moles of H2\mathrm{H_2}H2​ are

n(H2)=0.22022.4=0.0098214 moln(\mathrm{H_2}) = \frac{0.220}{22.4} = 0.0098214\,\text{mol}n(H2​)=22.40.220​=0.0098214mol

  1. Use stoichiometry to find moles of magnesium

Since Mg:H2=1:1\mathrm{Mg:H_2} = 1:1Mg:H2​=1:1,

n(Mg)=0.0098214 moln(\mathrm{Mg}) = 0.0098214\,\text{mol}n(Mg)=0.0098214mol

  1. Calculate mass of magnesium

m=n×Mm = n \times Mm=n×M

m(Mg)=0.0098214×24=0.2357 gm(\mathrm{Mg}) = 0.0098214 \times 24 = 0.2357\,\text{g}m(Mg)=0.0098214×24=0.2357g

  1. Convert to mg

0.2357 g=235.7 mg0.2357\,\text{g} = 235.7\,\text{mg}0.2357g=235.7mg

This is approximately:

236 mg236\,\text{mg}236mg

  1. Check options
  • A: 0.24 mg0.24\,\text{mg}0.24mg ❌
  • B: 235.7 g235.7\,\text{g}235.7g ❌
  • C: 2.444 g2.444\,\text{g}2.444g ❌
  • D: 236 mg236\,\text{mg}236mg ✅

Therefore, the correct option is D.

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