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 .

= 0.134; = 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  () = 1 - mole fraction of propanol ().

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

Furthermole, the partial pressure of isopropanol = *vapor pressure of isopropanol = 0.753*45.2 Torr = 34.04 Torr

The partial pressure of propanol = *vapor pressure of propanol = 0.247*20.9 Torr = 5.16 Torr

Similarly,

In the vapor phase,

The mole fraction of propanol () =

Where, is the partial pressure of propanol and is the partial pressure of isopropanol.

Therefore,

= 5.26/(34.04+5.16) = 0.134

= 1 - 0.134 = 0.866

answer; ///(u-235); ///is the correct

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