What is molar mass?

Question

The freezing point of an aqueous solution containing an unknown solute is -2.60 ∘C. The solution was prepared by dissolving 5.00 g of a nonelectrolytic solute in 100. mL of water. What is the molar mass of the unknown solute (in units of g mol^-1)?

*ΔT_f = K_fC_mi, K_f = -1.86 °C/m

Answer 1

Colligative properties:

Freezing point depression

∆T = imK

∆T = change in freezing point = 2.60ºC (b/c normal freezing point is 0ºC and now it is -2.60º)

i = van't Hoff factor = 1 because the solute is a non electrolyte

m = molality = moles of solute per kg of solvent = ?

K = freezing point constant for water = 1.86ºC/m

So we can now solve for m:

m = ∆T / (i)(K) = 2.60 / (1)(1.86) ... didn't use - sign for 2.60º and 1.86 as they simply produce a + sign

m = 1.398 moles of solute / kg of solvent

Now, how many kg of solvent do we have? We have 100. ml of water and assuming a density of 1g/ml we have 100 g water = 0.100 kg of water.

Now we calculate moles of solute:

1.398 moles / kg x 0.1 kg = 0.1398 moles

Since molar mass = grams / moles we now have all the info we need to find molar mass

grams = 5.00 gg

moles = 0.1398 moles

molar mass = 5.00 g / 0.1398 moles

molar mass = 35.8 g / mol