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

First, we need to calculate the mole fraction of each component in the mixture before any acetone is added.

For the butanone:
moles of butanone = (mass of butanone) / (molar mass of butanone)
moles of butanone = (1000 g) / (72.11 g/mol)
moles of butanone = 13.88 mol

For the acetone:
moles of acetone = (mass of acetone) / (molar mass of acetone)
moles of acetone = (x g) / (58.08 g/mol)

The total moles in the mixture is the sum of the moles of butanone and acetone:
total moles = moles of butanone + moles of acetone

The mole fraction of each component is:
mole fraction of butanone = moles of butanone / total moles
mole fraction of acetone = moles of acetone / total moles

We want the total pressure to be 125 mmHg, and we can use Dalton's law of partial pressures to relate the mole fractions to the partial pressures:

total pressure = pressure of butanone + pressure of acetone
125 mmHg = mole fraction of butanone * 100 mmHg + mole fraction of acetone * 232 mmHg

We have two equations (one for the number of moles and one for the partial pressures) and two unknowns (moles of acetone and mole fraction of acetone). We can solve for moles of acetone by rearranging the mole fraction equation:

mole fraction of acetone = 1 - mole fraction of butanone
moles of acetone / total moles = 1 - moles of butanone / total moles
moles of acetone = total moles - moles of butanone

Substituting into the partial pressure equation gives:

125 mmHg = (moles of butanone / total moles) * 100 mmHg + [(total moles - moles of butanone) / total moles] * 232 mmHg

Solving for total moles gives:

total moles = 17.59 mol

Then we can solve for the mass of acetone needed to reach this total moles:

moles of acetone = total moles - moles of butanone
moles of acetone = 17.59 mol - 13.88 mol
moles of acetone = 3.71 mol

mass of acetone = moles of acetone * molar mass of acetone
mass of acetone = 3.71 mol * 58.08 g/mol
mass of acetone = 215 g

Therefore, the amount of acetone that must be added is 215 g, which is closest to choice C (188 g).

Acetone (Mw = 58