A commercial bleach solution contains 3.62% by Mass of NaOCI in water,  A. Mole fraction of NaOCI in the sol. B. Molality of NaOCI in solut,?

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  • 4 weeks ago
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    A.

    Take a hypothetical sample of exactly 100 g of the bleach solution.

    (3.62 g NaOCl) / (74.4422 g NaOCI/mol) = 0.048628 mol NaOCl

    (100 - 3.62) g H2O / (18.01532 g H2O/mol) = 5.349891 mol H2O

    (0.048628 mol NaOCl) / (0.048628 mol + 5.349891 mol) =

    0.00901 (the mole fraction of NaOCl)

    B.

    (0.048628 mol NaOCl) / (100 - 3.62) g H2O x (1 kg/1000 g) = 0.50454 mol/kg =

    0.505 m NaOCl

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