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Some Basic Concepts of Chemistry question

2020 · 3 Sep · Shift 1 · Q9
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Some Basic Concepts of Chemistry question

2020 · 3 Sep · Shift 1 · Q9

JEE MainChemistrySome Basic Concepts of ChemistryNumerical+4 / −1
The volume strength of 8.9 M H2O2H_2O_2H2​O2​ solution calculated at 273 K and 1 atm is ‾\underline{\hspace{2cm}}​. (R = 0.0821 L atm K-1 mol-1) (rounded off ot the nearest integer)
Numerical answer
View written solutionFree

Correct answer: 100

  1. Meaning of volume strength

    Volume strength of a H2O2H_2O_2H2​O2​ solution means:

    Number of volumes of O2O_2O2​ (at STP) liberated by 1 volume of the solution.

  2. Decomposition reaction of hydrogen peroxide

    2H2O2→2H2O+O22H_2O_2 \rightarrow 2H_2O + O_22H2​O2​→2H2​O+O2​

    From the reaction:

    • 222 moles of H2O2H_2O_2H2​O2​ give 111 mole of O2O_2O2​
  3. Given molarity

    The solution is 8.9 M8.9\,M8.9M, so:

    1 L solution contains 8.9 mol H2O21\text{ L solution contains } 8.9\text{ mol } H_2O_21 L solution contains 8.9 mol H2​O2​

  4. Moles of oxygen produced from 1 L solution

    Using stoichiometry:

    moles of O2=8.92=4.45\text{moles of } O_2 = \frac{8.9}{2} = 4.45moles of O2​=28.9​=4.45

  5. Volume of O2O_2O2​ at 273 K273\,K273K and 1 atm1\,atm1atm

    Use ideal gas equation:

    V=nRTPV = \frac{nRT}{P}V=PnRT​

    V=(4.45)(0.0821)(273)1V = \frac{(4.45)(0.0821)(273)}{1}V=1(4.45)(0.0821)(273)​

    First calculate:

    0.0821×273=22.41330.0821 \times 273 = 22.41330.0821×273=22.4133

    Then:

    V=4.45×22.4133=99.74 LV = 4.45 \times 22.4133 = 99.74\text{ L}V=4.45×22.4133=99.74 L

  6. Volume strength

    Since this oxygen comes from 1 L of solution, volume strength is:

    99.7499.7499.74

    Rounded to nearest integer:

    100\boxed{100}100​

  7. Comparison with stored answer

    Stored correct answer = 100100100

    Our derived answer also is 100100100, so they agree.

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