Question

3 moles of liquid A and 1 mole of liquid B are mixed to form an ideal solution. The vapour pressure of solution becomes $500 \text{ mm Hg}$. If 1 mole of A is further added then vapour pressure of solution increases by $20 \text{ mm Hg}$. Find vapour pressure of pure B ($P_B^o$) in $\text{mm Hg}$ ?

• A. 200
• B. 400
• C. 600
• D. 800

Hint:

When solving systems of equations derived from Raoult's Law for multiple compositions, algebraic subtraction is the simplest way to find the vapor pressure of one pure component.

Answer 1

The Correct Option is A

Initial mixture: $n_A = 3$, $n_B = 1$. Mole fractions $X_A = 3/4$, $X_B = 1/4$. $P_S = 500 \text{ mm Hg}$.
Using Raoult's Law: $P_S = P_A^o X_A + P_B^o X_B$.
$500 = P_A^o (3/4) + P_B^o (1/4)$.
$2000 = 3P_A^o + P_B^o$ (1).

Second mixture: $n'_A = 4$, $n'_B = 1$. $X'_A = 4/5$, $X'_B = 1/5$. $P'_S = 500 + 20 = 520 \text{ mm Hg}$.
$520 = P_A^o (4/5) + P_B^o (1/5)$.
$2600 = 4P_A^o + P_B^o$ (2).

Subtract Equation (1) from Equation (2):
$(4P_A^o + P_B^o) - (3P_A^o + P_B^o) = 2600 - 2000$.
$P_A^o = 600 \text{ mm Hg}$.
Substitute $P_A^o$ into Equation (1):
$2000 = 3(600) + P_B^o$.
$P_B^o = 2000 - 1800 = 200 \text{ mm Hg}$.