A mixture is 10.0 mole% methyl alcohol, 75.0 mole% methyl acetate (C3H6O2), and 15.0 mole% acetic acid. Calculate the mass fractions of each compound. What is the average molecular weight of the mixture? What would be the mass (kg) of a sample containing 25.0 kmol of methyl acetate?

As we are given that a mixture is 10.0 mole% methyl alcohol, 75.0 mole% methyl acetate, and 15.0 mole% acetic acid that means 10.0 mole methyl alcohol, 75.0 mole methyl acetate, and 15.0 mole acetic acid present in 100 moles of mixture.

First we have to calculate the mole fraction of methyl alcohol, methyl acetate and acetic acid.

[tex]\text{Mole fraction of methyl alcohol}=\frac{10.0}{100}=0.100[/tex]

[tex]\text{Mole fraction of methyl acetate}=\frac{75.0}{100}=0.750[/tex]

[tex]\text{Mole fraction of acetic acid}=\frac{15.0}{100}=0.150[/tex]

Now we have to calculate the average molecular weight of the mixture.

Average molar mass = (Fraction of methyl alcohol × Molar mass of methyl alcohol) + (Fraction of methyl acetate × Molar mass of methyl acetate) + (Fraction of acetic acid × Molar mass of acetic acid)

Molar mass of methyl alcohol = 32.04 g/mol

Molar mass of methyl acetate = 74.08 g/mol

Molar mass of acetic acid = 60.05 g/mol

Average molar mass = (0.100 × 32.04) + (0.750 × 74.08) + (0.150 × 60.05)

Average molar mass = 67.8 g/mol

Now we have to calculate the mass (kg) of a sample containing 25.0 kmol of methyl acetate.

25 kmol methyl acetate = 25 × 1000 moles methyl acetate

As, 0.75 moles methyl acetate present in 1 mole compound

So, (25 × 1000) 0.75 moles methyl acetate present in compound = [(25×1000) × (1/0.75)] = 33333.33 mole

Now we have to calculate the mass of sample.

Mass of sample = Moles × Molar mass of compound

Mass of sample = 33333.33 mol × 67.8 g/mol

Mass of sample = 2259999.774 g = 2259.9 kg

Thus, the mass of sample will be, 2259.9 kg