An aqueous solution containing 35.9 g of an unknown molecular (nonelectrolyte) compound in 150.0 g of water has a freezing point of —1.3 °C. Calculate the molar mass of the unknown compound.

molr mass (mm):

mm = 342.43 g/mol

Explanation:

freezing point (ΔTf):

∴ ΔTf = - 1.3°C

∴ a: water (solvent)

∴ b: solute

∴ mb: molality of the solute [=] mol/Kg

∴ Kf: cryoscopic constant

⇒ Kf water = 1.86 °C.Kg/mol

∴ wb = 35.9 g

∴ wa = 150.0 g = 0.150 Kg

∴ molar mass solute (mm) [=] g/mol

⇒ mb = - ΔTf/Kf

⇒ mb = - ( - 1.3°C)/(1.86 °CKg/mol)

⇒ mb = 0.6989 mol/Kg

∴ moles solute (nb):

⇒ nb = (0.6989 mol/Kg)(0.150 Kg) = 0.1048 mol

molar mass:

⇒ mm = (35.9 g)/(0.1048 mol) = 342.43 g/mol

the answer is:  

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