1. Use Raoult’s law for an ideal binary solution

For liquids $A$ and $B$ behaving ideally,

$P_{\text{total}} = x_A P_A^0 + x_B P_B^0$

where:

• $x_A, x_B$ are mole fractions in liquid phase
• $P_A^0, P_B^0$ are vapour pressures of pure $A$ and pure $B$

Let:

$P_A^0 = a, \qquad P_B^0 = b$

2. Form equations from the two compositions

Case 1:

Given $x_A=0.7$, $x_B=0.3$, total pressure $=350$ mm Hg

$0.7a + 0.3b = 350$

Multiply by 10:

$7a + 3b = 3500 \qquad (1)$

Case 2:

Given $x_A=0.2$, $x_B=0.8$, total pressure $=410$ mm Hg

$0.2a + 0.8b = 410$

Multiply by 10:

$2a + 8b = 4100 \qquad (2)$

3. Solve the simultaneous equations

From (1):

$7a + 3b = 3500$

From (2):

$2a + 8b = 4100$

Eliminate $b$.

Multiply (1) by 8:

$56a + 24b = 28000$

Multiply (2) by 3:

$6a + 24b = 12300$

Subtract:

$50a = 15700$

$a = \frac{15700}{50} = 314$

Thus,

$P_A^0 = 314\ \text{mm Hg}$

4. Final answer

The vapour pressure of pure $A$ is:

$\boxed{314\ \text{mm Hg}}$

Nearest integer = $314$.

5. Comparison with stored answer

Stored correct answer = $314$

Our derived answer also equals $314$, so they agree.