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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.

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baraltoa

Answer:

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 ) )

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