Hello,

Vapour pressure of A, Va = 600 mmhg ( Given)

Vapour pressure of reaction, V = 760 mmhg ( Given)

No. of moles of A, Xa = No. of moles of B, Xb = 0.5 (Given)  [ Xa + Xb = 1]

Now applying Raoult's law, we have:

V = Va * Xa + Vb * Xb

We have to find, vapour pressure of B, Vb. Subsituting the values given, in the equation above, we get:

760 = 600 * 0.5 + Vb * 0.5

= Vb = 460/0.5 = 920

Therefore, Vapour pressure of B in pure state is: 920 mmhg (Answer)