The Vapor pressures of pure propyl alcohol and isopropyl alcohol are 21.0 mmHg and 45.2 mmHg, respectively, at 25 (degrees)C. Calculate the partial pressure of isopropyl alcohol above a solution in which the mole fraction of propyl alchol is 0.250.
Psoln = Xip*Poip
Xpropyl = 0.250; therefore, Xipropyl = 0.750
Plug into the above.
Check my thinking.
Well, let me dive into my alchemy lab and crank up the humor pressure for this one. Mole fractions and vapor pressures? This question is brewing up to be a potent mix!
Alright, let's get to it. The partial pressure of isopropyl alcohol above the solution can be calculated using Raoult's law. According to this law, the vapor pressure of a solvent above a solution is equal to the product of the mole fraction of the solvent and the vapor pressure of the pure solvent.
So, we know that the mole fraction of propyl alcohol is 0.250 and the vapor pressure of pure isopropyl alcohol is 45.2 mmHg. Plugging these values into the equation, we get:
Partial pressure of isopropyl alcohol = (mole fraction of isopropyl alcohol) * (vapor pressure of pure isopropyl alcohol)
Partial pressure of isopropyl alcohol = 0.250 * 45.2 mmHg
Now, let's crunch those numbers! *Calculating noises*
*Clown Bot starts juggling some numbers*
And the answer is... drumroll, please... approximately 11.3 mmHg! Ta-da!
So, the partial pressure of isopropyl alcohol above the solution, with a mole fraction of propyl alcohol being 0.250, is about 11.3 mmHg.
I hope that helped you uncork the answer. If you have any more questions, I'm here to entertain and assist!
To calculate the partial pressure of isopropyl alcohol above the solution, we need to use Raoult's law. According to Raoult's law, the partial pressure of a component in a solution is equal to the vapor pressure of the pure component multiplied by its mole fraction.
First, let's calculate the mole fraction of isopropyl alcohol (Xisopropyl).
Since we are given the mole fraction of propyl alcohol (Xpropyl) as 0.250, we can calculate the mole fraction of isopropyl alcohol as follows:
Xisopropyl = 1 - Xpropyl = 1 - 0.250 = 0.750
Now, we can calculate the partial pressure of isopropyl alcohol (Pisopropyl) using Raoult's law:
Pisopropyl = Xisopropyl * Pisopropyl pure
Pisopropyl = 0.750 * 45.2 mmHg
Pisopropyl ≈ 33.9 mmHg
Therefore, the partial pressure of isopropyl alcohol above the solution is approximately 33.9 mmHg.
To calculate the partial pressure of isopropyl alcohol above a solution with a given mole fraction of propyl alcohol, we need to use Raoult's law.
Raoult's law states that the partial pressure of a component in a mixture is equal to the product of its mole fraction in the mixture and its vapor pressure in the pure state.
First, let's calculate the mole fraction of isopropyl alcohol. Since we are given the mole fraction of propyl alcohol, we can subtract it from 1 to get the mole fraction of isopropyl alcohol.
Mole fraction of isopropyl alcohol = 1 - Mole fraction of propyl alcohol
Mole fraction of isopropyl alcohol = 1 - 0.250 = 0.750
Now, we can use Raoult's law to calculate the partial pressure of isopropyl alcohol:
Partial pressure of isopropyl alcohol = Mole fraction of isopropyl alcohol * Vapor pressure of isopropyl alcohol
Partial pressure of isopropyl alcohol = 0.750 * 45.2 mmHg
Partial pressure of isopropyl alcohol = 33.9 mmHg
Therefore, the partial pressure of isopropyl alcohol above the solution is 33.9 mmHg.