Question

1‑propanol ( n ‑propanol) and 2‑propanol (isopropanol) form ideal solutions in all proportions. Calculate the partial pressure and the mole fraction ( y ) of the vapor phase of each component in equilibrium with each of the given solutions at 25 °C. P ∘ prop = 20.9 Torr and P ∘ iso = 45.2 Torr at 25 °C. A solution with a mole fraction of x prop = 0.247 .

Answer 1

Answer:

$y_{prop}$ = 0.134; $y_{iso}$ = 0.866

The partial pressure of isopropanol = 34.04 Torr; The partial pressure of propanol = 5.26 Torr

Explanation:

For each of the solutions:

mole fraction of isopropanol  ($x_{iso}$) = 1 - mole fraction of propanol ($x_{prop}$).

Given: mole fraction of propanol = 0.247. Thus, the mole fraction of isopropanol = 1 - 0.247 = 0.753.

Furthermole, the partial pressure of isopropanol = $x_{iso}$*vapor pressure of isopropanol = 0.753*45.2 Torr = 34.04 Torr

The partial pressure of propanol = $x_{prop}$*vapor pressure of propanol = 0.247*20.9 Torr = 5.16 Torr

Similarly,

In the vapor phase,

The mole fraction of propanol ($y_{prop}$) = $\frac{P_{prop} }{P_{prop}+P_{iso}}$

Where, $P_{prop}$ is the partial pressure of propanol and $P_{iso}$ is the partial pressure of isopropanol.

Therefore,

$y_{prop}$ = 5.26/(34.04+5.16) = 0.134

$y_{iso}$ = 1 - 0.134 = 0.866