Answer:
Partial pressure is 0.13 atm
Explanation:
CHECK THE COMPLETE QUESTION BELOW :
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.
Using Raoult´s law for ideal soultions we have
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
Raoult's law states that the vapor pressure of a solution is dependent on the mole fraction of a solute added to the solution.
Raoult's law can be expressed below
Psolution = ΧsolventP0solvent.
Expressing it interns of the constituents given in the question we have
P(CH₃OH) = X(CH₃OH) x Pº(CH₃OH)
To calculate the mole fraction of CH₃OH, we make use of the formula below :
X(A) = mol (A) / ntotal
Ntotal = (sum of number of moles of A )+( moles solvent)
mol (CH₃3OH) can be calculated as :: 88.2 g/ 32 g/mol = 2.76 mol of CH₃OH
mol (H₂O) can be calculated as ::116 g/ 18 g/mol = 6.44 mol
total n = (6.44 + 2.76) mol = 9.20 mol
To calculate the partial pressure the we say;
P(CH₃OH) = (2.76 mol CH + 9.20 mol) x( 0.43 atm) = 0.13 atm
Hence, the partial pressure rounded to two significant figures is 0.13 atm
