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Some Basic Concepts of Chemistry question

2004 · Shift 0 · Q52
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Some Basic Concepts of Chemistry question

2004 · Shift 0 · Q52

JEE MainChemistrySome Basic Concepts of ChemistryMCQ+4 / −1
6.02 ×\times× 1020 molecules of urea are present in 100 ml of its solution. The concentration of urea solution is (Avogadro constant, NA = 6.02 ×\times× 1023 mol-1)
  1. A
    0.02 M
  2. B
    0.01 M
  3. C
    0.001 M
  4. D
    0.1 M
View written solutionFree

Correct answer: B

  1. Given data

    • Number of urea molecules =6.02×1020= 6.02 \times 10^{20}=6.02×1020
    • Volume of solution =100 mL=0.1 L= 100\,\text{mL} = 0.1\,\text{L}=100mL=0.1L
    • Avogadro constant NA=6.02×1023 mol−1N_A = 6.02 \times 10^{23}\,\text{mol}^{-1}NA​=6.02×1023mol−1

  2. Calculate moles of urea

    Using moles=number of moleculesNA\text{moles} = \frac{\text{number of molecules}}{N_A}moles=NA​number of molecules​

    n=6.02×10206.02×1023=10−3 mol=0.001 moln = \frac{6.02 \times 10^{20}}{6.02 \times 10^{23}} = 10^{-3}\,\text{mol} = 0.001\,\text{mol}n=6.02×10236.02×1020​=10−3mol=0.001mol

  3. Calculate molarity

    Molarity is given by M=nVM = \frac{n}{V}M=Vn​

    where VVV is in litres.

    M=0.0010.1=0.01 mol L−1M = \frac{0.001}{0.1} = 0.01\,\text{mol L}^{-1}M=0.10.001​=0.01mol L−1

  4. Identify the correct option 0.01 M\boxed{0.01\,M}0.01M​

    So, the correct option is B.

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