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Some Basic Concepts of Chemistry question

2020 · 9 Jan · Shift 2 · Q6
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Some Basic Concepts of Chemistry question

2020 · 9 Jan · Shift 2 · Q6

JEE MainChemistrySome Basic Concepts of ChemistryNumerical+4 / −1
10.30 mg of O2O_2O2​ is dissolved into a liter of sea water of density 1.03 g/mL. The concentration of O2O_2O2​ in ppm is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 10

  1. Understand ppm in this context

For dilute solutions by mass, ppm=mass of solutemass of solution×106\text{ppm} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 10^6ppm=mass of solutionmass of solute​×106

  1. Given data
  • Mass of dissolved O2=10.30 mgO_2 = 10.30\,\text{mg}O2​=10.30mg
  • Volume of sea water =1 L=1000 mL= 1\,\text{L} = 1000\,\text{mL}=1L=1000mL
  • Density of sea water =1.03 g mL−1= 1.03\,\text{g mL}^{-1}=1.03g mL−1
  1. Find mass of 1 L sea water

mass=ρV=1.03×1000=1030 g\text{mass} = \rho V = 1.03 \times 1000 = 1030\,\text{g}mass=ρV=1.03×1000=1030g

Convert into mg: 1030 g=1030×1000=1.03×106 mg1030\,\text{g} = 1030 \times 1000 = 1.03 \times 10^6\,\text{mg}1030g=1030×1000=1.03×106mg

  1. Calculate ppm

ppm=10.301.03×106×106\text{ppm} = \frac{10.30}{1.03 \times 10^6} \times 10^6ppm=1.03×10610.30​×106

ppm=10.301.03=10\text{ppm} = \frac{10.30}{1.03} = 10ppm=1.0310.30​=10

  1. Final answer

10\boxed{10}10​

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