A mixture of 0.2mole of alcohol of A and 0.5 mole of alcohol B has a total vapour pressure of 40mmHg at 298K. if the mixture obeys Raoult's law, find the pure vapour pressure of B at 298K given that the pure vapour pressure of A is 20mmHg at 298K
VP total (mix) = 40mmHg(given)
VP(gas) = mole fraction of gas x total pressure of mix
mole fraction of X(B) = 0.5/(0.2 + 0.5) = 0.7142
VP(B) = X(B) x TTL Pressure = 0.7142(40mmHg) = 28.568mmHg ~ 30mmHg (1 sig.fig)
To find the pure vapor pressure of alcohol B, we can use Raoult's law, which states that the vapor pressure of a component in an ideal mixture is directly proportional to the mole fraction of that component in the mixture.
First, let's determine the mole fractions of each alcohol in the mixture:
Mole fraction of alcohol A (Xa) = moles of A / total moles
Xa = 0.2 moles / (0.2 moles + 0.5 moles) = 0.2 / 0.7 = 2/7
Mole fraction of alcohol B (Xb) = moles of B / total moles
Xb = 0.5 moles / (0.2 moles + 0.5 moles) = 0.5 / 0.7 = 5/7
Now, we can use Raoult's law to find the partial pressure of alcohol B (Pb) in the mixture:
Pb = Xb * pure vapor pressure of B
Pb = (5/7) * pure vapor pressure of B
Given that the total vapor pressure of the mixture is 40 mmHg and the pure vapor pressure of A is 20 mmHg, we can express the total vapor pressure as the sum of the partial pressures of the two components:
Total vapor pressure = Pa + Pb
40 mmHg = 20 mmHg + Pb
Simplifying the equation, we get:
Pb = 40 mmHg - 20 mmHg
Pb = 20 mmHg
So, the pure vapor pressure of alcohol B is 20 mmHg at 298K.