The problem asks for the mole percentage of component 'D' in a binary solution of 'A' and 'D' at its boiling point. We are given the vapor pressures of pure liquids 'A' and 'D', and the total pressure at which the solution boils.

Applying Raoult's and Dalton's Laws

For an ideal binary solution, the total vapor pressure ($P_{total}$) is related to the mole fractions of the components ($X_A$, $X_D$) and their respective pure vapor pressures ($P_A^\circ$, $P_D^\circ$) by Raoult's Law and Dalton's Law:

According to Raoult's Law:

• Partial pressure of A ($P_A$) = $X_A \cdot P_A^\circ$
• Partial pressure of D ($P_D$) = $X_D \cdot P_D^\circ$

According to Dalton's Law:

• Total pressure ($P_{total}$) = $P_A + P_D$

Combining these:

$P_{total} = (X_A \cdot P_A^\circ) + (X_D \cdot P_D^\circ)$

Since the sum of mole fractions is 1, we have $X_A = 1 - X_D$. Substituting this into the equation:

$P_{total} = ((1 - X_D) \cdot P_A^\circ) + (X_D \cdot P_D^\circ)$

$P_{total} = P_A^\circ - (X_D \cdot P_A^\circ) + (X_D \cdot P_D^\circ)$

$P_{total} = P_A^\circ + X_D (P_D^\circ - P_A^\circ)$

Calculating Mole Fraction of D

We can rearrange the equation to solve for $X_D$:

$X_D (P_D^\circ - P_A^\circ) = P_{total} - P_A^\circ$

$X_D = \frac{P_{total} - P_A^\circ}{P_D^\circ - P_A^\circ}$

Substituting Given Values

Given values are:

• $P_A^\circ = 500$ mm Hg
• $P_D^\circ = 800$ mm Hg
• $P_{total} = 700$ mm Hg (at the boiling point of the solution)

Substitute these values into the formula for $X_D$:

$X_D = \frac{700 \text{ mm Hg} - 500 \text{ mm Hg}}{800 \text{ mm Hg} - 500 \text{ mm Hg}}$

$X_D = \frac{200 \text{ mm Hg}}{300 \text{ mm Hg}}$

$X_D = \frac{2}{3}$

Determining Mole Percentage

To find the mole percentage of 'D', we multiply the mole fraction by 100%:

Mole percentage of D = $X_D \times 100\%$

Mole percentage of D = $\frac{2}{3} \times 100\% = 66.67\%$