How many grams are 3.01 * 1023 molecules of CuSO4?

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Nikita

Chemistry

14 April, 15:15

How many grams are 3.01 * 1023 molecules of CuSO4?

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Answers (1)

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Anthony Ballard · Answer 1

Answer is: 79.8 grams of copper (II) sulfate.

N (CuSO₄) = 3.01·10²³; number of molecules.

n (CuSO₄) = N (CuSO₄) : Na.

n (CuSO₄) = 3.01·10²³ : 6.02·10²³ 1/mol.

n (CuSO₄) = 0.5 mol; amount of substance.

m (CuSO₄) = n (CuSO₄) · M (CuSO₄).

m (CuSO₄) = 0.5 mol · 159.6 g/mol.

m (CuSO₄) = 79.8 g; mass of substance.

M - molar mass.