3.29. A mixture of methanol (methyl alcohol) and water contains 60.0% water by m

Question

3.29. A mixture of methanol (methyl alcohol) and water contains 60.0% water by mass. (a) Assuming volume additivity of the components, estimate the specific gravity of the mixture at 20°C. What volume (in liters) of this mixture is required to provide 150 mol of methanol? (b) Repeat Part (a) with the additional information that the specific gravity of the mixture at 20°C is 0.9345 (making it unnecessary to assume volume additivity). What percentage error results from the volume-additivity assumption?

Explanation / Answer

Part a

Let's take the initial mass of mixture = 100 g

Mass of water = 60 g

Density of water at 20°C = 0.9982 g/cm3

Volume of water = mass/density

= 60g / 0.9982 g/cm3

= 60.1082 cm3

Mass of methanol = 40 g

Density of methanol at 20°C = 0.7917 g/cm3

Volume of methanol = mass/density

= 40g / 0.7917 g/cm3

= 50.5241 cm3

Based on volume additivity

Total volume = 60.1082 + 50.5241 = 110.6323 cm3

Density of mixture = mass of mixture /volume of mixture

= 100/110.6323

= 0.9038 g/cm3

Specific gravity of mixture = 0.9038

Mass of given methanol = moles x molecular weight

= 150 mol x 32.04 g/mol

= 4806 g

40 g methanol consists = 110.6323 cm3 of mixture

4806 g methanol consists =( 110.6323 x 4806/40) cm3 of mixture

= 13292.47 cm3 x 10^-3L/cm3

= 13.292 L

Part b

If specific gravity of mixture = 0.9345

Density of mixture = 0.9345 g/cm3

Volume of mixture = mass of mixture / density of mixture

= 100/0.9345 = 107.0091 cm3

40 g methanol consists = 107.0091 cm3 of mixture

4806 g methanol consists =( 107.0091 x 4806/40) cm3 of mixture

= 12857.14 cm3 x 10^-3L/cm3

= 12.857 L

Percent error in volume

= (ideal value - actual value) * 100 / (actual value)

= (13.292 - 12.857)*100/12.857

= 3.383%

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