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Does anyone know how to do this problem, i can't seem to solve it. If 39.25 g of a substance is dissolved in 155.5 g of benzene solvent, the freezing point of the solution is 0.08 oC. Calculate the apparent molar mass (g/mol) of substance.

If 1030 g of a potassium citrate is dissolved in 2320 g of water solvent, calculate the freezing temperature (if below 0 oC, include the sign) of the solution. Consider if the solute is an electrolyte. Kf (oC/m) 1.858

so this is what i did, 0.08C = (5.065Ckg/mol)(x/0.1555) which is 0.002456071. i then took 39.25g/0.002456071 which is 15980 g/mol. This seems to big for molar mass of a substance

is the answer 5.38??? this assignement is due in less than 20 min and i need an answer