Amixture is 20.00 mole% methyl alcohol, 60.0 mole% methyl acetate, and 20.0 mole% acetic acid.
what is the mass of a sample containing 45.0 kmol of methyl acetate?

4714.950 kilograms  is the mass of a sample containing 45.0 kmol of methyl acetate.

Explanation:

Moles of methyl acetate ==45.0 kmol= 45000 mol

Mole percentage of methyl acetate = 60.0%

Total moles in the sample = n

Mole percentage of methyl alcohol = 20.0%

Moles of methyl alcohol = n_2

Mass of methyl alcohol =  

=

Mole percentage of acetic acid  = 20.0%

Moles of acetic acid = n_3

Mass of acetic acid=

Mass of methyl methyl acetate=

Mass of sample: 480,600 g + 3,333,600 g + 900,750 g = 4714950 g

4714950 g = 4714.950 kg

(1 kg = 1000 g)

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