Molar mass with boiling point?

Question

A solution of 35.7 g of a nonelectrolyte in 220.0 g of chloroform has a boiling point of 64.5 °C. Assuming ideal solution behavior, what is the molar mass of this compound?

Answer 1

For boiling point elevation problems, we use the formula ∆T = imK. Unfortunately, you did not provide the normal boiling point for chloroform, but looking it up, we find it to be around 61.2ºC

∆T = change in boiling point = 64.5º - 61.2º = 3.3º

i = van't Hoff factor = 1 for a non electrolyte

m = molality = moles of non electrolyte / kg chloroform = ?

K = boiling point constant for chloroform = 3.88ºKg/m (looked it up on Wikipedia)

Plugging in the values and solving for m, we have...

3.3 = (1)(m)(3.88)

m = 0.851 moles of unknown / kg

Since we don't have 1 kg of chloroform, but rather we have only 220.0 g = 0.2200 kg, we can find the moles of unknown.

moles of unknown = 0.851 mol/kg x 0.2200 kg = 0.187 moles

Now that we have the moles, and we know we started with 35.7 g of the unknown, it is a simple matter to determine the molar mass of the unknown.

Molar mass = 35.7 g / 0.187 moles = 191 g/mole.

Answer 2

Hello Evanly,

To answer this question you have to use boiling point elevation formula: ΔTb​= i * Kb * ​m, where i is the van't Hoff factor, which is 1 in this case, since the solute is not an electrolyte, Kb is the boiling point elevation constant for chloroform (3.63 kg*C/mol), m is the molality (mol solute/kg solvent), and ΔTb difference in boiling point between pure chloroform (61.2°C) and the solution at hand (64.5°C).

If you plug in the variables, you can solve for molality to be m = 3.3°C/3.63 kg*C/mol = 0.91 mol/kg. Then you take the molality and find the moles of the solute by multiplying 0.22 kg of chloroform, you should get 0.2 moles. Lastly, you find the molar mass by dividing the 0.2 moles of the solute by 35.7 g of the solute, which should yield 178 g/mol as the final answer.

Hope this helps,

Danil