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2 March, 21:28

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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  1. 2 March, 22:06
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    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.
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