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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?

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2 Answers

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

ΔTf = -Kf*Cm*i

You've got values for

ΔTf = -2.5°C

Kf = 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 (Cm) 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.

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.