SHOW YOUR WORK. gnoring vaporization, what volumes of ethanol and water are need
ID: 548906 • Letter: S
Question
SHOW YOUR WORK. gnoring vaporization, what volumes of ethanol and water are needed to prepare 250.0mL of a fraction of ethanol equal to 0.090? Under these particular binary solution that has a mole conditions, the partial molar volumes and densities are (52.2 mL mol and 0.7893 g mL1) and (18.1 ml mol' and 0.997 g mL') for ethanol and water, respectively. "You know, it would have been faster just to calculate rather than measure the vapor pressure up here." 0.0452 is 92.0 It is found that the standard boiling point of a binary solution of A and B with x °C. At this temperature, the vapor pressures of A and B above the mixture are 0.750 bar and 0.250 bar, respectively. The Henry's Law constant for B is 25.0 bar. The activity coefficient for A is 1.19. You may assume the vapour behaves ideally. What are the activity and activity coefficient for B? 2. What is the vapour pressure of pure A? The following temperature-composition data were obtained for a mixture of two liquids A and B at 1.00 bar: 3. 125 0.91 0.99 130 0.65 0.91 135 0.45 0.77 140 0.30 0.61 145 0.18 0.45 150 0.098 0.25 XA The boiling points for pure A and pure B are 124 °C and 155 °C, (a) Plot the temperature- composition diagram for the mixture (b) What is the composition of the vapor in equilibrium with the liquid when xA -0.25? When xe (9) What is the standard boiling point of an A/B mixture that is initially x-0,402 W = 0.55? compositi point? on of the first drop of vapor, and what is the partial pressure of A at that boiling hat is theExplanation / Answer
Answer for Ques. No-1
Given in ques.,
Total volume of binary solution = 250 mL
Mole fraction of ethanol = 0.09
To find,
vol. of water in 250 mL sol. = ?
vol. of ethanol in 250 mL sol. = ?
Solution :-
Let the volume of water = x mL
then volume of ethanol = 250 -x mL
no. of moles of water = x mL/ 18.1 mL mol-1 = Xwater ----------------------------------------------(1)
no. of moles of ethanolr = (250-x) mL / 52.5 mL mol-1 = Xethanol -----------------------------------(2)
mole fraction of water (Y1) = Xwater / (Xwater + Xethanol) = 0.91
mole fraction of ethanol (Y2 ) = Xethanol / (Xwater + Xethanol) = 0.09 [as Y1 + Y2 = 1 ]
Now,
Y2 / Y1 = Xethanol / Xwater = 0.09 / 0.91 = 0.1
From (1) and (2),
Xethanol / Xwater = {(250-x) mL / 52.5 mL mol-1 } / {x mL/ 18.1 mL mol-1} = 0.1
=> {(250-x) mL / 52.5 mL mol-1 } / {x mL/ 18.1 mL mol-1} = 0.1
=> (250-x) * 18.1 = 52.5 * x * 0.1
=> 4525 - 18.1x = 5.25x
=> 4525 = 23.35x
=> x = 4525 / 23.35 = 193.8
In the begining of solution, vol. of water was assumed as x mL
and ethanol was 250 - x
Thus,
vol. of water in 250 mL binary sol. = 193.8 mL
vol. of water in 250 mL binary sol. = 56.2 mL
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