How many grams are in 4.00 x 1023 particles of alcl3?

Question

Triston Evans

Chemistry

14 October, 21:39

How many grams are in 4.00 x 1023 particles of alcl3?

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Answers (1)

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Bree · Answer 1

Answer is: 88,9 g of AlCl₃.

N (AlCl₃) = 4,00 · 10²³.

n (AlCl₃) = N (AlCl₃) : Na.

n (AlCl₃) = 4 · 10²³ : 6·10²³ 1/mol

n (AlCl₃) = 0,66 mol.

m (AlCl₃) = n (AlCl₃) · M (AlCl₃)

m (AlCl₃) = 0,66 mol · 133,35 g/mol = 88,9 g.

Na - Avogadro number.

M - molar mass.

n - amount of substance