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

If the solution of liquids $A$ and $B$ boils at $80^\circ\text{C}$ under $1\,\text{atm}$ pressure, then its total vapour pressure at $80^\circ\text{C}$ must be equal to external pressure:

$P_{\text{total}} = 760\ \text{mm Hg}$

For an ideal binary mixture:

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

where

• $P_A^0 = 520\ \text{mm Hg}$
• $P_B^0 = 1000\ \text{mm Hg}$
• $x_A + x_B = 1$

2. Substitute $x_B = 1 - x_A$

$760 = x_A(520) + (1-x_A)(1000)$

3. Solve for $x_A$

$760 = 520x_A + 1000 - 1000x_A$

$760 = 1000 - 480x_A$

$480x_A = 1000 - 760 = 240$

$x_A = \frac{240}{480} = 0.5$

4. Convert to mol percent

$x_A = 0.5 = 50\%$

So, the amount of $A$ in the mixture is:

$\boxed{50\ \text{mol percent}}$

5. Check options

• A: $52\%$ ❌
• B: $34\%$ ❌
• C: $48\%$ ❌
• D: $50\%$ ✅

Therefore, the correct option is D.