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

If the gas vapor mix is 0.2 mol A and 0.5 mole B where the total pressure is 40-mmHg, The partial pressure of A will NOT be 20-mmHg. It's more like 11.428-mmHg and B will be 28.572-mmHg.

0.2 mole A + 0.5 mole B => 0.7 mol mix.
mole fraction A = (0.2/0.7) = 0.2857
mole fraction B = (0.5/0.7) = 0.7143
∑mole fractions = 0.2857 + 0.7143 = 1.000

VP(A) = X₁∙VP₁ = 0.2857(40-mmHg) = 11.428-mmHg
VP(B) = X₂∙VP₂ = 0.7143(40-mmHg) = 28.572-mmHg

To find the vapor pressure of alcohol B at 298K, we can use Raoult's law. According to Raoult's law, the vapor pressure of a component in a mixture is proportional to its mole fraction in the mixture.

Let's calculate the mole fraction of alcohol B in the mixture first.

Mole fraction (X) of alcohol B = Moles of B / Total moles

Moles of A = 0.2
Moles of B = 0.5

Total moles in the mixture = Moles of A + Moles of B = 0.2 + 0.5 = 0.7

Mole fraction of alcohol B = 0.5 / 0.7 = 0.714

According to Raoult's law, the vapor pressure of B (P_B) can be calculated as:

P_B = X_B * P°_B

Where:
X_B = Mole fraction of B in the mixture
P°_B = Pure vapor pressure of B

We know that the vapor pressure of A (P_A) is 20 mmHg, so we can use this information to solve for P°_B.

P_A = X_A * P°_A

Where:
X_A = Mole fraction of A in the mixture
P°_A = Pure vapor pressure of A

Rearranging the equation, we can solve for P°_A:

P°_A = P_A / X_A

Substituting the given values:

P°_A = 20 mmHg / (1 - X_B)
P°_A = 20 mmHg / (1 - 0.714)
P°_A = 20 mmHg / 0.286
P°_A ≈ 69.93 mmHg

Now we can calculate the pure vapor pressure of B:

P_B = X_B * P°_B

Substituting the values calculated earlier:

P_B = 0.714 * 69.93 mmHg
P_B ≈ 49.993 mmHg

Therefore, the pure vapor pressure of alcohol B at 298K is approximately 49.993 mmHg.