A mixture of 0.2mole of alcohol A and 0.5mole of alcohol has a total vapour pressure of 40mmHg at 298k,if the mixture obey's Raoult's law,find the pure pressure of B at 298k given that the pressure of A is 20mmHg at 298k
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To find the pure pressure of alcohol B at 298K, given that the pressure of alcohol A is 20mmHg at 298K, we need to use Raoult's law and consider the mole fraction of each component in the mixture.
According to Raoult's law, the vapor pressure of a component in a mixture is equal to the product of its mole fraction and its pure vapor pressure. Mathematically, it can be expressed as:
P(A) = X(A) * P(A)°
P(B) = X(B) * P(B)°
Where:
P(A) is the vapor pressure of component A in the mixture
P(B) is the vapor pressure of component B in the mixture
X(A) is the mole fraction of component A in the mixture
X(B) is the mole fraction of component B in the mixture
P(A)° is the pure vapor pressure of component A
P(B)° is the pure vapor pressure of component B
In this case, we are given:
P(A) = 20mmHg
P(A)° = 40mmHg (total vapor pressure of the mixture)
X(A) = 0.2 moles of A / (0.2 moles of A + 0.5 moles of B)
X(B) = 0.5 moles of B / (0.2 moles of A + 0.5 moles of B)
First, we can determine the mole fraction of A and B in the mixture:
X(A) = 0.2 / (0.2 + 0.5) = 0.2 / 0.7 = 0.2857
X(B) = 0.5 / (0.2 + 0.5) = 0.5 / 0.7 = 0.7143
Using Raoult's law, we can calculate the pure vapor pressure of B:
P(B) = X(B) * P(B)°
= 0.7143 * 40mmHg
≈ 28.57mmHg
Therefore, the pure pressure of alcohol B at 298K is approximately 28.57mmHg.
XA = mols A/total mols = ? = estimated 0.3
XB = mols B/total mols = estimated 0.7
Ptotal = XA*PoA + XB*PoB.
You know Ptotal, XA, XB, and PoA. Solve for PoB.
Post your work if you get stuck.