1. We are interested in increasing the concentration of a SL methanol-water mixt

Question

1. We are interested in increasing the concentration of a SL methanol-water mixture that contains 30% by volume of methanol. The liquid mixture which is initially at 20°C is heated to 65°C in the still, which is connected to other units as shown. Initial pressure is at atmospheric condition. The entire assembly is completely sealed so that no vapors escape. Then vapors are condensed in a condenser by continuously circulating large amount of cold water at 25°C. Liquid is collected as a condensate. If 1.OL of liquid with 31.2% (molar basis) of methanol is present as a condensate after 1 h, what will be the composition of the remaining liquid in the still? What volume of liquid will be remaining in the still? er of water. fructose Heater & stirrer

Explanation / Answer

Basis : 100 mol of condensate

31.2% methanol and rest is water ( 68.8%).

Density of methanol = 729 g/L ; Molar mass of methanol = 32 g/mol

Density of water = 1000 g/l ; Molar mass of methanol = 18 g/mol

Moles of Methanol = Mole fraction * Total moles = 0.312*100 = 31.2 moles

Mass of methanol = Moles * molar mass = 31.2 * 32g/mol = 998.4 g

Volume of methanol = Mass / density = 998.4/729 = 1.369 L

Mass of water = Moles of water * molar mass = 68.8 * 18 = 1238.4 g

Volume of water = Mass /density = 1238.4 / 1000 = 1.238 L

Total volume of the mixture = 1.369 + 1.238 = 2.607 L

Total mass of the mixture = 998.4 + 1238.4 = 2236.8

Density of mixture = 2236.8/2.607 = 857.99 g/L

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Now we see that,100 mol solution with 31.2 mol% methanol = 2.607 L

1 L of solution= 100/2.607 = 38.358 molesof solution

Moles of methanol in it = 38.358 * 0.312 = 11.967

Mass of methanol = Moles of methanol * molar mass = 11.967 * 32 = 382.944 g

Moles of water in 38.358 moles of solution = 0.688*38.358 = 26.39 mol

Mass of water = Moles of water * molar mass = 26.39 *18 = 475.02 g

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In feed side i.e still ,

Volume of methanol = 0.3*5 = 1.5 L

Mass of methanol = 1.5 * 729 = 1093.5 g

Volume of water = 0.7*5 = 3.5 L

Mass of water = 3.5 * 1000 = 3500 g

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After distillation ,

Mass of methanol in feed = Mass of methanol at the start of operation - mass of methnol collected as condensate = 1093.5 - 382.944 = 710.556 g

Volume of methanol = Mass of methanol/ Density of methanol = 710.556/729 = 0.974 L

Mass of water = Mass of water at the start of operation - mass of water collected as condensate = 3500 - 475.02 = 3024.98 g

Volume of water = 3024.98/1000 = 3.025 L

Volume fraction of methanol in the still = Volume of methanol / Total Volume = 0.974 / ( 0.974 +3.025) = 0.243

Volume fraction of water = 1 - 0.243 = 0.757

Total volume remaining in th still = 0.974 + 3.025 = 3.999 L  

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