Butanone, CH3CH2COCH3, has a vapor pressure of 100. Torr at 25°C; at the same temperature, propanone, CH3COCH3, has a vapor pressure of 222 Torr. What mass of propanone must be mixed with 322.0 g of butanone to give a solution with a vapor pressure of 135 Torr? Assume ideal behavior.
Use Psoln = XsolventPsolvento
That will calculate X for each solvent.
Go from there to grams of each (one is given in grams). Post your work if you get stuck.
To determine the mass of propanone needed to produce a solution with a vapor pressure of 135 Torr, we can use Raoult's law, which states that the vapor pressure of a component in an ideal solution is directly proportional to its mole fraction.
The mole fraction (X) of a component can be calculated by dividing the number of moles of that component by the total number of moles in the solution.
First, let's calculate the number of moles of butanone (CH3CH2COCH3) in the solution. We know that the molar mass of butanone is 58.08 g/mol.
Number of moles of butanone = mass of butanone / molar mass of butanone
= 322.0 g / 58.08 g/mol
= 5.54 mol
Next, let's calculate the mole fraction of propanone (CH3COCH3) needed in the solution, assuming the total number of moles remains constant.
Mole fraction of propanone = moles of propanone / (moles of butanone + moles of propanone)
= moles of propanone / (5.54 mol + moles of propanone)
Now, we can use Raoult's law to set up an equation using the vapor pressures of butanone, propanone, and the vapor pressure of the solution:
Vapor pressure of solution = (mol fraction of butanone) x (vapor pressure of butanone) + (mol fraction of propanone) x (vapor pressure of propanone)
135 Torr = (1 - mol fraction of propanone) x 100 Torr + (mol fraction of propanone) x 222 Torr
Simplifying the equation, we get:
135 Torr = 100 Torr + (mol fraction of propanone) x (222 Torr - 100 Torr)
35 Torr = (mol fraction of propanone) x 122 Torr
mol fraction of propanone = 35 Torr / 122 Torr
mol fraction of propanone ≈ 0.287
Now, we'll use this mole fraction to calculate the number of moles of propanone in the solution:
0.287 = moles of propanone / (5.54 mol + moles of propanone)
Rearranging the equation, we have:
0.287 (5.54 mol + moles of propanone) = moles of propanone
Expanding and isolating the moles of propanone:
1.59 mol + 0.287 moles of propanone = moles of propanone
0.713 moles of propanone = moles of propanone
Finally, we can calculate the mass of propanone needed:
Mass of propanone = moles of propanone x molar mass of propanone
= 0.713 mol x 58.08 g/mol
≈ 41.4 g
Therefore, approximately 41.4 grams of propanone must be mixed with 322.0 grams of butanone to give a solution with a vapor pressure of 135 Torr.