Question
Asked Jan 16, 2019
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A mixture of 20 mL of isoamyl acetate (MW=130.2 g/mol and density= 0.88 g/mL) and 20 mL of methyl benzoate (MW= 136.2 g/mol and density =1.09 g/mL) is distilled. Calculate the mole percent for each component. Use these mole percents and the figure below to answer the following questions.
a. What is the initial boiling point of this mixture?
b. What is the composition of the vapor in equilibrium with the liquid? Is this composition the same as the composition of the initial condensate from simple distillation?
c. Instead of a simple distillation, you decide to use fractional distillation. Assuming two theoretical plates, what is the composition of the first fraction collected?

200
200
190
190
180
180
Vapor
170
Temp oC
170
Liquid
160
160
150
150
140
140
50
25
Mole % Isoamyl Acetate . 100
Mole % Methyl Benzoate
75
50
75
100
25
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200 200 190 190 180 180 Vapor 170 Temp oC 170 Liquid 160 160 150 150 140 140 50 25 Mole % Isoamyl Acetate . 100 Mole % Methyl Benzoate 75 50 75 100 25

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Expert Answer

Step 1

Volume of isoamyl acetate mixture = 20 mL

Molecular weight of isoamyl acetate = 130.2 g/mol 

Density of isoamyl acetate = 0.88 g/mL

Volume of methyl benzoate mixture = 20 mL

Molecular weight of methyl benzoate = 136.2 g/mol 

Density of methyl benzoate = 1.09 g/mL

Density is equal to the ratio of mass to the volume. The mathematical expression is given as: 

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Step 2

Number of moles is equal to the ratio of given mass to the molar mass of the substance. 

First, calculate the number of moles of isoamyl acetate and methyl benzoate.

To calculate the number of moles, first calculate the mass of compound.

Mass of isoamyl acetate and methyl benzoate is calculated as:

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Step 3

Now, calculate the number of moles.

Number of moles of isoam...

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