How many grams of ch3oh must be added to water to prepare 325 ml of a solution that is 4.50 mch3oh?

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Arthur

Chemistry

29 December, 02:47

How many grams of ch3oh must be added to water to prepare 325 ml of a solution that is 4.50 mch3oh?

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Deon Hahn · Answer 1

Answer is: 46,8 grams of methanol must be added.

V (solution) = 325 ml : 1000 ml/l = 0,325 l.

c (CH₃OH) = 4,5 M = 4,5 mol/L.

n (CH₃OH) = V (solution) · c (CH₃OH).

n (CH₃OH) = 0,325 l · 4,5 mol/l.

n (CH₃OH) = 1,4625 mol.

m (CH₃OH) = n (CH₃OH) · M (CH₃OH).

m (CH₃OH) = 1,4625 mol · 32 g/mol.

m (CH₃OH) = 46,8 g.