Two volatile liquids X and Y form an ideal solution at 298 K and have vapour pressures equal to 100 mm and 200 mm of Hg respectively in their pure state. The mole fraction of X in the solution is 0.4 and the mole fraction of Y in the vapour phase is a/20. Calculate the value of a.

Given the pure vapour pressures $P^{\circ}_{X} = 100$ mm Hg and $P^{\circ}_{Y} = 200$ mm Hg.

The mole fraction of X in the liquid phase is $x_{X} = 0.4$. Therefore, the mole fraction of Y in the liquid phase is $x_{Y} = 1 - 0.4 = 0.6$.

Using Raoult's Law, the partial pressures of X and Y are:
$P_{X} = x_{X} P^{\circ}_{X} = 0.4 \times 100 = 40$ mm Hg
$P_{Y} = x_{Y} P^{\circ}_{Y} = 0.6 \times 200 = 120$ mm Hg

The total pressure of the solution is $P_{total} = P_{X} + P_{Y} = 40 + 120 = 160$ mm Hg.

The mole fraction of Y in the vapour phase, denoted as $y_{Y}$, is given by:
$y_{Y} = \dfrac{P_{Y}}{P_{total}} = \dfrac{120}{160} = \dfrac{12}{16} = \dfrac{3}{4} = 0.75$.

Given that $y_{Y} = \dfrac{a}{20}$, we equate the values:
$\dfrac{a}{20} = \dfrac{3}{4}$

$a = \dfrac{3}{4} \times 20 = 3 \times 5 = 15$.

Answer: $15$