When 5g of a noneelectrolyte is added to 25g of water, the new freezing point is -2.5 C. What is the molecular mass of the unknown compound?

Question

How do I start this problem?

Answer 1

deltaT = imK

where deltaT is the change in temperature = 2.5 degrees C

i = van't Hoff factor = 1 for a non electrolyte since it doesn't ionize

m = molality = moles solute/kg solvent = ?

K = freezing point depression constant for water = 1.853

Rearranging you have m = deltaT/(i)(K) = 2.5/(1)(1.853) = 1.35 molal

This means there are 1.35 moles of the solute in 1 kg of solvent.  The problem tells you that you have 5 grams of solute in 25 grams of solute (0.025 kg solvent), so you can now set up a relationship to determine how many grams of solute there are in 1 Kg of solvent and that will be equal to the 1.35 moles determined above.  You do this as follows:

5 grams/0.025 Kg = x grams/1 Kg  and x = 200 grams in 1 Kg which makes 1.35 molal.  This means that 200 grams is equal to 1.35 moles, and since molar mass = grams/mole you have 200 grams/1.35 moles = 148 g/mol and so this would be the molecular mass or molar mass of the unknown.

Answer 2

Since you listed this as basic chemistry I'm going to assume that you're using the simplified equation

ΔT_f = -K_f*C_m*i

You've got values for

ΔT_f = -2.5°C

K_f = 1.853 (K*kg)/mol

i = 1 because it is a non-electrolyte, there's no ionization

The only other trick would be to remember that molal concentration (C_m) is moles of solute per kg of solvent, and then that moles of solute is equal to

5g solute*(1 mol/x g solute).

so you end up with

2.5 = 1.853*[(5/x)/(0.025 kg)]

Just solve for x.