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kondaur [170]
4 years ago
5

Calculate the molarity of a solution that contains 3.00 grams Na2SO4 in 25 mL of solution.

Chemistry
1 answer:
Naddik [55]4 years ago
3 0

Answer:

M Na2SO4 sln = 0.8448 M

Explanation:

  • molarity (M) [=] mol/L

∴ mass Na2SO4 = 3.00 g

∴ volume soln = 25 mL = 0.025 L

∴ molar mass Na2SO4 = 142.04 g/mol

⇒ mol Na2SO4 = (3.00 g)*(mol/142.04 g) = 0.02112 mol

⇒ M Na2SO4 sln = (0.02112 mol/0.025 L ) = 0.8448 M

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S = 32

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3.- Convert these masses to moles

                               27 g of Al ----------------- 1 mol

                               31.5 g ----------------------  x

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                               16 g of O ----------------  1 mol

                               56.1 g of O -------------  x

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                               32 g of S ---------------  1 mol

                               12.4 g of S -------------   x

                                x = 0.39 moles

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S  =   0.39 / 0.39 = 1

5.- Write the empirical equation

                                Al₃O₉S

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