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Some Basic Concepts of Chemistry question

2024 · 31 Jan · Shift 2 · Q29
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Some Basic Concepts of Chemistry question

2024 · 31 Jan · Shift 2 · Q29

JEE MainChemistrySome Basic Concepts of ChemistryNumerical+4 / −1
The molarity of 1 L1 \mathrm{~L}1 L orthophosphoric acid (H3PO4)\left(\mathrm{H}_3 \mathrm{PO}_4\right)(H3​PO4​) having 70%70 \%70% purity by weight (specific gravity 1.54 g cm−31.54 \mathrm{~g} \mathrm{~cm}^{-3}1.54 g cm−3) is ‾M\underline{\hspace{2cm}}\mathrm{M}​M. (Molar mass of H3PO4=98 g mol−1\mathrm{H}_3 \mathrm{PO}_4=98 \mathrm{~g} \mathrm{~mol}^{-1}H3​PO4​=98 g mol−1)
Numerical answer
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Correct answer: 11

  1. Given data

    • Volume of solution =1 L=1000 cm3= 1\,\text{L} = 1000\,\text{cm}^3=1L=1000cm3
    • Specific gravity =1.54 g cm−3= 1.54\,\text{g cm}^{-3}=1.54g cm−3
    • Purity by weight =70%= 70\%=70%
    • Molar mass of H3PO4=98 g mol−1\mathrm{H_3PO_4} = 98\,\text{g mol}^{-1}H3​PO4​=98g mol−1

  2. Mass of 1 L solution Using density: Mass=ρV=1.54×1000=1540 g\text{Mass} = \rho V = 1.54 \times 1000 = 1540\,\text{g}Mass=ρV=1.54×1000=1540g

  3. Mass of pure H3PO4\mathrm{H_3PO_4}H3​PO4​ in the solution Since the acid is 70%70\%70% pure by weight: Mass of H3PO4=0.70×1540=1078 g\text{Mass of } \mathrm{H_3PO_4} = 0.70 \times 1540 = 1078\,\text{g}Mass of H3​PO4​=0.70×1540=1078g

  4. Number of moles of H3PO4\mathrm{H_3PO_4}H3​PO4​ n=107898=11 moln = \frac{1078}{98} = 11\,\text{mol}n=981078​=11mol

  5. Molarity Molarity is moles of solute per litre of solution: M=111=11 MM = \frac{11}{1} = 11\,\text{M}M=111​=11M

  6. Final answer 11\boxed{11}11​

The derived answer matches the stored correct answer.

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