Vapour pressures of pure liquids ’A’ and ’D’ at 50◦C are 500 mm Hg and 800 mm Hg respectively. The binary solution of ’A’ and ’D’ boils at 50◦C and 700 mm Hgpressure. The mole percentage of ’D’ in the solution is:

The correct option is 66.67 mole percent.

Step-by-Step Explanation:

We are given the following data for a binary solution of two pure liquids, 'A' and 'D', at a temperature of 50°C:
- Vapour pressure of pure liquid A, $P_A^{\circ }=500\text{mm Hg}$
- Vapour pressure of pure liquid D, $P_D^{\circ }=800\text{mm Hg}$
- Total vapour pressure of the boiling solution, $P_{\text{total}}=700\text{mm Hg}$

According to Raoult's Law, the total vapour pressure ($P_{\text{total}}$) of a binary solution containing components A and D is given by the formula:

$P_{\text{total}}=P_A^{\circ }{x}_A+P_D^{\circ }{x}_D$

where $x_A$ and $x_D$ represent the mole fractions of A and D in the liquid solution respectively.

Since the sum of the mole fractions in a binary mixture is always equal to 1:
$x_A+x_D=1$
We can express $x_A$ in terms of $x_D$ as:
$x_A=1-x_D$

Substituting this relation into the Raoult's Law equation gives:

$P_{\text{total}}=P_A^{\circ }(1-x_D)+P_D^{\circ }{x}_D$

Now, substitute the given numerical values into the equation:

$700=500(1-x_D)+800x_D$

Expand and simplify the equation to solve for $x_D$:
$700=500-500x_D+800x_D$
$700=500+300x_D$
Subtract 500 from both sides:
$200=300x_D$
Divide by 300:

$x_D=\frac{200}{300}=\frac{2}{3}\approx 0.6667$

To convert the mole fraction of D to mole percentage, we multiply by 100:
Mole percentage of D = $x_D\times 100=0.6667\times 100=66.67\text{mole percent}$