Acetone (mw = 58.08, P =232 mmHg) and butanone (mw = 72.11, P
=100 mmHg) have the indicated molar masses and vapor pressures. A container holds 1.00 kg of butanone. How much acetone must be added to the butanone to elevate the total vapor pressure over the mixture to 125 mmHg at 25°C?
A) 5.313 kg
B) 290 g
C) 188 g
D) More information needed
E) None of the above

Question

Acetone (mw = 58.08, P  =232 mmHg) and butanone (mw = 72.11, P 
=100 mmHg) have the indicated molar masses and vapor pressures. A container holds 1.00 kg of butanone. How much acetone must be added to the butanone to elevate the total vapor pressure over the mixture to 125 mmHg at 25°C?
A) 5.313 kg 
B) 290 g 
C) 188 g 
D) More information needed 
E) None of the above

Quizplus · Accepted Answer

The total vapor pressure of the mixture can be calculated using Raoult's Law:

PTotal = PAcetone x XAcetone + PButanone x XButanone

where PTotal is the total vapor pressure of the mixture, PAcetone and PButanone are the vapor pressures of acetone and butanone respectively, and XAcetone and XButanone are the mole fractions of acetone and butanone in the mixture.

Since the mass of butanone is given, we can calculate its corresponding mole fraction:

nButanone = mButanone / MButanone = 1.00 kg / 72.11 g/mol = 13.87 mol
nTotal = nAcetone + nButanone
XButanone = nButanone / nTotal = 13.87 mol / (13.87 mol + nAcetone)

Now we can rewrite Raoult's Law as:

PTotal = PAcetone x XAcetone + PButanone x (1 - XAcetone)

We are given that PTotal = 125 mmHg, PButanone = 100 mmHg, and PAcetone = 232 mmHg. Substituting these values into the equation and solving for XAcetone, we get:

125 = 232 XAcetone + 100 (1 - XAcetone)
XAcetone = 0.191

This means that the mole fraction of acetone in the mixture needs to be 0.191. Since we know the mass of butanone and the total mass of the mixture (1.00 kg + mass of acetone), we can use this mole fraction to find the mass of acetone needed:

XAcetone = nAcetone / nTotal
nAcetone = XAcetone x nTotal
mAcetone = nAcetone x MAcetone
mTotal = mAcetone + mButanone = 1.00 kg + mAcetone

Substituting the mole fraction and solving for the mass of acetone, we get:

0.191 = nAcetone / (nAcetone + 13.87)
nAcetone = 2.91 mol
mAcetone = nAcetone x MAcetone = 2.91 mol x 58.08 g/mol = 169.3 g

Therefore, the mass of acetone needed to raise the total vapor pressure to 125 mmHg is 169.3 g or 0.1693 kg, which is closest to answer choice (C).

Acetone (Mw = 58