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Some Basic Concepts of Chemistry question

2022 · 26 Jun · Shift 1 · Q1
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Some Basic Concepts of Chemistry question

2022 · 26 Jun · Shift 1 · Q1

JEE MainChemistrySome Basic Concepts of ChemistryMCQ+4 / −1
A commercially sold conc. HCl is 35% HCl by mass. If the density of this commercial acid is 1.46 g/mL, the molarity of this solution is: (Atomic mass : Cl = 35.5 amu, H = 1 amu)
  1. A
    10.2 M
  2. B
    12.5 M
  3. C
    14.0 M
  4. D
    18.2 M
View written solutionFree

Correct answer: C

  1. Given data

    • Commercial HCl is 35% by mass
    • Density of solution =1.46 g mL−1= 1.46\,\text{g mL}^{-1}=1.46g mL−1
    • Molar mass of HCl =1+35.5=36.5 g mol−1= 1 + 35.5 = 36.5\,\text{g mol}^{-1}=1+35.5=36.5g mol−1

  2. Take 1 litre of solution Since molarity is moles per litre, consider 1000 mL1000\,\text{mL}1000mL of solution.

    Mass of 111 L solution: 1000×1.46=1460 g1000 \times 1.46 = 1460\,\text{g}1000×1.46=1460g

  3. Find mass of HCl in this solution The solution is 35%35\%35% HCl by mass, so mass of HCl in 1460 g1460\,\text{g}1460g solution is: 35100×1460=511 g\frac{35}{100} \times 1460 = 511\,\text{g}10035​×1460=511g

  4. Convert mass of HCl into moles Moles of HCl=51136.5=14.0\text{Moles of HCl} = \frac{511}{36.5} = 14.0Moles of HCl=36.5511​=14.0

  5. Calculate molarity Since these are moles in 111 litre of solution, M=14.0 mol L−1=14.0 MM = 14.0\,\text{mol L}^{-1} = 14.0\,\text{M}M=14.0mol L−1=14.0M

  6. Option check

    • A: 10.2 M10.2\,\text{M}10.2M ❌
    • B: 12.5 M12.5\,\text{M}12.5M ❌
    • C: 14.0 M14.0\,\text{M}14.0M ✅
    • D: 18.2 M18.2\,\text{M}18.2M ❌

Therefore, the correct answer is Option C.

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