We assume that this gas is an ideal gas. We use the ideal gas equation to calculate the amount of the gas in moles. It is expressed as:
PV = nRT
(672) (1/760) (36.52) = n (0.08206) ( 68 +273.15)
n = 1.15 mol of gas
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Answer:
= 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
True I think is the answer
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