Q. On mixing, heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components (heptanes and octane) are 105 kPa and 45kPa respectively. Vapour pressure of the solution obtained by mixing 25 g of heptanes and 35 g of octane will be (molar mass of heptanes = 100 g ${mol}^{- 1}$ and of octane = 114 g${mol}^{- 1}$).
A) 72.0 kPa
B) 36.1 kPa
C) 96.2 kPa
D) 144.5 kPa
Explanation
(a)
$P_{T} = X_{H}.\ P_{H}^{\circ}\ + \ P_{O}^{\circ}\$
$X_{H} = \frac{\frac{25}{100}}{\frac{25}{100} + \frac{35}{114}}\ = 0.45$ $and\ \therefore\ \ \ X_{o} = 0.55$
$P_{T}\ = \ 0.45\ \times 105 + 0.55 \times 45 = 72kPa$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.