Problem 2 (Mixtures - 30 points): A mixture of ethanol and water contains 40% wa

Question

Problem 2 (Mixtures - 30 points): A mixture of ethanol and water contains 40% water by mass. a) [10 Points) Assuming volume additivity of the components, estimate the specific gravity of the mixture at 20°C. b) [10 Points) Based on this assumption, what volume (in liters) of this mixture is required to provide 200 gmol of ethanol? [10 Points) Repeat parts a) and b) with the additional information that the specific gravity of the mixture at 20°C is 0.93518 (making it unnecessary to assume volume additivity). What percentage error results from the volume additivity assumption? c) Some potentially useful information: SG AÁ. (T.) AA,(T) Compound Formula Mol. Wt. (20°/4°) T(°C) kJ/mol T.(C) kJ/mol T.(K), Pe(atm) Ethyl alcohol CH3OH 46.070.789 -114.6 5.021 78.5 38.58 516.3 63.0 (Ethanol) Water H,0 18.016 1.000.00 6.0095 100.00 40.656 647.4 218.3 (AA) kJ/mol -277.63(1) -235.31 (g) - 285.84(1) - 241.83(g) (AA). kJ/mol -1366.91(1) - 1409.25(8) You can assume that the density of water is the same at 20°C as it is at 4°C.

Explanation / Answer

Part a

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

Mass of water = 40 g

Density of water at 20°C = 1 g/cm3

Volume of water = mass/density

= 40g / 1 g/cm3

= 40 cm3

Mass of ethanol = 60 g

Density of ethanol at 20°C = 0.789 g/cm3

Volume of ethanol = mass/density

= 60g / 0.789 g/cm3

= 76.0456 cm3

Based on volume additivity

Total volume = 40 + 76.0456 = 116.0456 cm3

Density of mixture = mass of mixture /volume of mixture

= 100/116.0456

= 0.8617 g/cm3

Specific gravity of mixture = 0.8617

Part b

Mass of given ethanol = moles x molecular weight

= 200 mol x 46 g/mol

= 9200 g

60 g methanol consists = 116.0456 cm3 of mixture

9200 g methanol consists =( 116.0456 x 9200/60) cm3 of mixture

= 17793.658 cm3 x 10^-3L/cm3

= 17.793 L

Part c

If specific gravity of mixture = 0.93518

Density of mixture = 0.93518 g/cm3

Volume of mixture = mass of mixture / density of mixture

= 100/0.93518 = 106.9312 cm3

60 g methanol consists = 106.9312 cm3 of mixture

9200 g methanol consists =( 106.9312 x 9200/60) cm3 of mixture

= 16396.13 cm3 x 10^-3L/cm3

= 16.396 L

Percent error in volume

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

= (17.793 - 16.396)*100/16.396

= 8.52%

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