Question

An unknown alcohol is analyzed by freezing point depression. The unknown is either methanol (CH3OH), ethanol (C2H5OH), or propanol, (C3H7OH). A solution made by adding 10.423 g of the unknown to 100.0mL of water freezes at -4.2 degrees C [water kFP= -1.853 degrees C.kg/mol, assume d(water)= 1g/mL]. Explain how you used the data to identify the alcohol.

Answer 1

mass of unknown = 10.423 g

mass of water = 100 g

delta Tf = Kf x m

0 - (- 4.2) = 1.853 x m

m = 2.266

molality = 2.266 m

molality = moles of solute / mass of solvent

2.266 = moles / 0.1

moles = 0.2266

moles = mass / molar mass

0.2266 = 10.423 / molar mass

molar mass = 45.985 g/mol

corresponding alcohol is Ethanol. because its molar mass - 46 g/mol