Two liquids A and B on mixing form an ideal solution. The vapour pressure of the solution containing 2 moles of A and 4 moles of B is 0.2 atm. If the vapour pressure of pure A at a certain temperature is 0.3 atm then what will be the vapour pressure of pure B at that temperature?
nA = 2 mols
nB = 4 mols
nTotal = 6 mols
XA = (2/6) = about 0.3
XB = (4/6) = about 0.7--you can do these more accurately and I've estimated the other values below, also, so you will need to redo the whole problem.
pA = XA*PoA
pA = 0.3* 0.3 = about 0.09
Ptotal = 0.2 = pA + pB
You know Ptotal and pA, solve for pB
Then pB = XB*PoB
You know pB and XB, solve for PoB.
To find the vapor pressure of pure B at the given temperature, we can use Raoult's law. According to Raoult's law, the vapor pressure of an individual component in an ideal solution is directly proportional to its mole fraction in the solution.
In this case, we have a solution containing 2 moles of A and 4 moles of B. The total moles of the solution is 2 + 4 = 6 moles.
To determine the mole fraction of B in the solution, we divide the number of moles of B by the total number of moles of the solution:
Mole fraction of B = 4 moles of B / 6 moles of solution = 2/3
Since the solution is ideal, the vapor pressure of the solution is directly proportional to the mole fraction of B. Given that the vapor pressure of the solution is 0.2 atm, we can set up the following equation:
Vapor pressure of B / Vapor pressure of the solution = Mole fraction of B
Vapor pressure of B / 0.2 atm = 2/3
Simplifying the equation:
Vapor pressure of B = (2/3) * 0.2 atm
Vapor pressure of B = 0.133 atm
Therefore, the vapor pressure of pure B at the given temperature is 0.133 atm.