Vapour pressure of pure liquids 'A' and 'D' at \(50^oC\) are \(500\, \ mm\) Hg and \(800\, \ mm\) Hg respectively. The binary solution of 'A' and 'D' boils at \(50^oC\) and \(700\, \ mm\)Hg pressure. The mole percentage of 'D' in the solution is

66.67 mole percent

The correct answer is option 2. 66.67 mole percent.

Given,

\(p^o_A = 500\, \ mm\, \ Hg\)

\(p^o_D = 800\, \ mm\, \ Hg\)

\(p_{total} = 700\, \ mm\, \ Hg\)

We know, from Raoult's law

\(p_A = p^o_A\chi _A\)

\(p_D = p^o_D\chi _D\, \ = p^o_D(1 - \chi _A)\)

Therefore, total pressure,

\(p_{total} = p_A + p_D\)

\(⇒ p_{total} = p^o_A\chi _A + p^o_D(1 - \chi _A)\)

\(⇒ p_{total} = p^o_A\chi _A + p^o_D - p^o_D\chi _A\)

\(⇒ p_{total} = (p^o_A - p^o_D)\chi _A + p^o_D\)

\(⇒ 700 = (500 - 800)\chi _A + 800\)

\(⇒ -100 = -300\chi _A\)

\(\chi _A = \frac{100}{300}\)

\(⇒ \chi _A = 0.3333\)

Therefore,

\(\chi _D = 1 - \chi _A\)

\(⇒ \chi _D = 1 - 0.3333\)

\(⇒ \chi _D = 0.6667\)

Therefore,

\(\% \chi _D = 0.6667 × 100\)

or, \(\% \chi _D = 66.67\text{ mole percent}\)