Question
Question: A mixture of ethyl alcohol and propyl alcohol has a vapor pressure of \[290{\text{ }}mm\] at \[300{\...
A mixture of ethyl alcohol and propyl alcohol has a vapor pressure of 290 mm at 300 K. The vapor pressure of propyl alcohol is 200 mm . If the mole fraction of ethyl alcohol is 0.6 its vapor pressure (in mm) at the same temperature will be:
A) 350
B) 300
C) 700
D) 360
Solution
In the question a mixture of two type of alcohol at 290 mmvapor pressure at a temperature of 300 Kare given and also the vapor pressure of one alcohol is given along with its vapor pressure. So we have to find the temperature of the ethyl alcohol at 0.6 vapor pressure.
Formula used: PT=PA∘×XA+PB∘×XB
Where, PT = total pressure
PA∘ = vapor pressure of component A
XA = mole fraction of component A in solution
PB∘ = vapor pressure of component B
XB = mole fraction of component B in solution
Complete step-by-step solution:
By using Raoult's law we will find the temperature of the alcohol. We know that when substances were mixed together in a solution, the vapor pressure of the solution decreased simultaneously.
Meanwhile, we know that, according to Raoult’s law
The total vapor pressure of binary mixture of miscible liquids be having ideally is given by
PT=PA∘×XA+PB∘×XB
Where XA and XB are mole fraction of A and B in the liquid phase
Whereas PA∘and PB∘ are vapor pressures of pure liquids.
By putting the values in the formula, we get:
⇒P=PA∘×XA+PB∘×XB
⇒290=PA∘×0.6 + 200×( 1−0.6 )
⇒290=0.6×PA∘+80
∴PA∘=350
Hence, the correct answer is option ‘A’.
Note: We can define Rault’s laws that when a solvent’s partial vapour pressure in a solution (or mixture) is equal or identical to the vapors pressure of the pure solvent multiplied by its mole fraction in the solution.