A mixture of 0.2 moles of alcohol A and 0.5 moles of alcohol B has a total vapour pressure of 40 mmHg at 297 K. If the mixture obeys Raoult's law, find the vapour pressure of pure B at 297 K given that the pressure of a pure A is 20 mmHg at 297 K.
Calculate X for A(which I'm calling a) and X for B (which I'm calling b). Poa and Pob are Poa and Pob respectively.
pa + pb = 40 mm
pa = Xa*Poa
Knowing pa you calculate pb
Then pb = Xb*Pob
You know pb and Xb, solve for pob.
Post your work if you get stuck and I can help you through it.
To find the vapor pressure of pure B at 297 K, we can first understand Raoult's law, which states that the partial pressure of a component in a mixture is proportional to its mole fraction.
In this case, we have a mixture of alcohol A and alcohol B. Let's denote the mole fraction of alcohol A as xA and the mole fraction of alcohol B as xB.
According to Raoult's law, the partial pressure of alcohol A (PA) in the mixture is given by:
PA = xA * P°A
where P°A is the vapor pressure of pure A at 297 K (given as 20 mmHg).
Similarly, the partial pressure of alcohol B (PB) in the mixture is given by:
PB = xB * P°B
We are trying to find the vapor pressure of pure B (P°B) at 297 K. So, let's rearrange the equation for PB:
P°B = PB / xB
Now, we need to determine the values of xA and xB. The mole fraction of a component is given by the number of moles of that component divided by the total number of moles in the mixture.
Given that we have 0.2 moles of alcohol A and 0.5 moles of alcohol B in the mixture, the total number of moles in the mixture is:
Total moles = 0.2 moles + 0.5 moles = 0.7 moles
Therefore, we can calculate the mole fractions as follows:
xA = moles of A / Total moles = 0.2 moles / 0.7 moles ≈ 0.286
xB = moles of B / Total moles = 0.5 moles / 0.7 moles ≈ 0.714
Substituting these values back into the equation, we can now calculate the vapor pressure of pure B (P°B):
P°B = PB / xB = 40 mmHg / 0.714 ≈ 56.02 mmHg
Therefore, the vapor pressure of pure B at 297 K is approximately 56.02 mmHg.