Question

An aqueous solution containing 34.2g of an unknown molecular (nonelectrolyte) compound in 143.1g of water was found to have a freezing point of -1.5?C. Calculate the molar mass of the unknown compound.

Answer 1

Molar Mass = grams / moles

                  = 34.2 grams / X moles

Change in Freezing point temperature

               = Tsolution - Tpure

               = -1.5^oC - 0^oC = -1.5^oC

Calculate the Molality , using the change in Boiling point and Elevation constant

        DT = mKf

   -1.5^oC = m * (-1.86^oC/m)

          m = -1.5^oC / -1.86^oC/m

          m   = 0.806m or 0.806 mol/kg

Find the moles of solute from molality by multiplying by the kg of solvent

    143.1 gr of water * 1kg/1000g = 0.143kg water

(0.806 mol/kg) * 0.143kg = 0.115 moles of solute

molar mass = grams / moles

    = 34.2 grams / 0.115 moles

molar mass of unknown compound is ,    = 297.4 g/mol