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

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

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