Question

An aqueous solution containing 17.5 g of an unknown molecular compound in 100.0 g of water was found to have a freezing point of -1.8

Answer 1

(change in freezing point) = (molal freezing point depression constant)(effective molality of solution)

1.8C = (1.86 C/m)(molality)
molality = 0.9677

molality = (moles solute) / (kg solvent)
0.9677 = (moles solute) / 0.100 kg
moles solute = 0.09677

molar mass = 17.5g / 0.09677 mol = 180.8 g/mol

Answer 2

mass of solute (w2) = 17.5 g

weight of solvent (water)(w1) = 100g

K_f for water = 1.86K Kg/mol

M = 1000 x K_f x w2/ $\Delta$ T_f x w1

= 1000 x 1.86 x 17.5 / 1.8 x 100 = 32550/180= 180.83 g/mol