A mixture of ethanol (C 2 H 5 OH) and water is analyzed and found to contain 30.
ID: 714288 • Letter: A
Question
A mixture of ethanol (C2H5OH) and water is analyzed and found to contain 30.0 mol% water. Assume that the specific gravity of a mixture of the two components varies linearly with the composition, i.e.,
SGmix = 0.79 + 0.21x
where x is the mass fraction water.
If 5 liters of pure water are added to the initial 5 liters of mixture, determine the volume of the resulting mixture.
How many liters of pure water must be added to 5 liters of the initial mixture to increase the total volume to exactly 25 liters?
A densitometer is used to measure the density of the ethanol-water mixture as it is diluted. If the measured density is 0.9956 g/cm3, how much pure water (L) has been added to the initial 5 L?
Explanation / Answer
1. Lets assume 100 moles of mixture is present.
Moles of water = 30
Mass of water = 30x18 = 540 g
Moles of Ethanol = 70
Mass of ethanol = 70 x 46 = 3220 g
Total mass of mixture = 540 + 3220 = 3760 g
Mass % of water = 540 / 3760 x 100
Mass % of water = 14.362 %
Mass of water added = 5 x 1000 = 5000 g
Total mass of water in mixture = 540 + 5000 = 5540 g
Total mass of mixture = 5540 + 3760 = 9300 g
Mass fraction of water = 5540 / 9300 = 0.596
SG = 0.79 + (0.21) x 0.596 = 0.915
Volume of mixture = 9300 / 0.915x1000 = 10.162 lit
2. Let Mass of water added = X x 1000 = 1000X g
Total mass of water in mixture = 540 + 1000X
Total mass of mixture = 540 + 3760 + 1000X = 4300 + 1000X g
Mass fraction of water = 540 + 1000X / 4300 + 1000X
SG = 0.79 + (0.21) x (540 + 1000X / 4300 + 1000X)
Volume of mixture = 9300 / (0.79 + (0.21) x (540 + 1000X / 4300 + 1000X)) x1000 = 25
Solve for X and that will be the lit of water.
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