What is the mass (in grams) of 9.93 ă— 1024 molecules of methanol (ch3oh)?

Answer is: mass of methanol is 528.32 grams.

N(CH₃OH) = 9.93·10²⁴; number of methanol molecules.

n(CH₃OH) = N(CH₃OH) ÷ Na (Avogadro constant).

n(CH₃OH) = 9.93·10²⁴ ÷ 6.022·10²³ 1/mol.

n(CH₃OH) = 16.49 mol; amount of substance.

m(CH₃OH) = n(CH₃OH) · M(CH₃OH).

m(CH₃OH) = 16.49 mol · 32.04 g/mol.

m(CH₃OH) = 528.32 g.

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