At a certain temperature, the vapor pressure of pure methanol is measured to be 0.43atm. Suppose a solution is prepared by mixing 88.2 g of methanol and 116. g of water. Calculate the partial pressure of methanol vapor above this solution. Be sure your answer has the correct number of significant digits. Note for advanced students: you may assume the solution is ideal.

0.13 atm

Explanation:

Our strategy here is to make use of Raoult's law for ideal soultions:

P (A) = X (A) Pº (A)

where P (A) is the partial vapor pressure pressure of methanol,

X (A) is the mole fraction of solute (methanol) in solution,

Pº (A) is the vapor pressure of pure solute

Then

P (CH₃OH) = X (CH₃OH) x Pº (CH₃OH)

We do not have the mole fraction of CH₃OH, but it could be calculated from the formula:

X (A) = mol (A) / total n

total n is the number of moles of A + moles solvent

mol (CH₃3OH) = 88.2 g / 32 g/mol = 2.76 mol CH₃OH

mol (H₂O) = 116 g / 18 g/mol = 6.44 mol

total n = (6.44 + 2.76) mol = 9.20 mol

Now we are in position to calculate the partial pressure asked:

P (CH₃OH) = (2.76 mol CH + 9.20 mol) x 0.43 atm = 0.13 atm

(rounded to two significant figures of the least precise number, 0.43 atm))