1‑propanol (nn ‑propanol) and 2‑propanol (isopropanol) form ideal solutions in all proportions. Calculate the partial pressure and the mole fraction (yy) of the vapor phase of each component in equilibrium with each of the given solutions at 25 °C. P∘prop=20.9 TorrPprop°=20.9 Torr and P∘iso=45.2 TorrPiso°=45.2 Torr at 25 °C. A solution with a mole fraction of xprop=0.243xprop=0.243.

Piso = 32.17 Torr

Pprop = 5.079 Torr

yprop = 0.1364

yiso = 0.8636

Explanation:

From the question; we can opine that:

The mole fraction of isopropanol in a mixture of isopropanol and propanol will be 1. The partial pressure of isopropanol will be its mole fraction multiplied by vapor pressure of isopropanol The partial pressure of propanol will be its mole fraction multiplied by vapor pressure of propanol In the vapor, the mole fraction of propanol will be its partial pressure divided by the sum of the two partial pressures

NOW;

When xprop = 0.243; xisopropanol will be 1 - 0.243 = 0.757

P°iso = 45.2 Torr at 25 °C so

Piso will be 45.2 * 0.757 = 32.17 Torr

Pprop will be 20.9 * 0.243 = 5.079 Torr

yprop = 5.079 / (5.079 + 32.17) = 0.1364

yiso = 1-0.1364 = 0.8636