A solution is prepared from 26.7 g of an unknown compound and 116.2 g acetone, C3H6O, at 313 K. At this temperature the vapor pressure of pure acetone is 0.526 atm while that of the solution is 0.501 atm. What is the molecular weight of the unknown compound?

The molecular weight of the unknown compound is 267.7 g/mol

Explanation:

Lowering vapor pressure → Colligative property where the vapor pressure of solution is lower than vapor pressure of pure solvent

ΔP = P° . Xm

0.526 atm - 0.501 atm = 0.025atm

0.025 atm = 0.526 atm . Xm

Xm = 0.025 atm / 0.526 atm → 0.0475 (mole fraction)

Mole fraction = Moles of solute / Total moles (solute + solvent)

0.0475 = Moles of solute / Moles of solute + Moles of solvent

We determine the moles of solvent → 116.2 g . 1mol / 58 g = 2 moles

0.0475 = Moles of solute / Moles of solute + 2

0.0475 moles of solute + 0.095 = Moles of solute

0.095 = Moles of solute - 0.0475 moles of solute

0.095 / 0.9525 = Moles of solute → 0.0997 moles

Molas mass = g/mol → 26.7 g / 0.0997 mol = 267.7 g/mol

a is the correct answer

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