Question

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 .

Answer 1

Piso = 32.17 Torr

Pprop = 5.079 Torr

yprop = 0.1364

yiso = 0.8636

Step-by-step 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