How many grams of sulfur (S8) would have to be dissolved in 125.0 g of carbon tetrachloride to lower the freez

Question

How many grams of sulfur (S8) would have to be dissolved in 125.0 g of carbon tetrachloride to lower the freezing point by 2.10 degrees celsius?

kf = 29.8 K-kg/mol for CCl4

Answer 1

As I said when you previously asked this question….

The freezing point depression of a solution can be found as:
ΔT = molality * k_f
Where molality is the concentration of the solute (in this case Sulfur) and k_f is the freezing point depression constant for the solvent.

You are given k_f in the question. You are told that the freezing point is lowered in the solution by 2.10 °C. All you need to calculate now is the concentration of the sulfur.

Rearranging the above equation to solve for the concentration,
Molality = ΔT / k_f
Which you should be able to easily plug in your known values and solve.

But we were asked to find the mass of Sulfur needed, not just its concentration. We know that,
Molality = moles of solute / kilograms of solvent
We are given the mass of the solvent (in terms of gram, which we will need to concert to kilograms). We already know what the concentration of the solution should be. All we need to do now if find the number of moles of Sulfur needed to achieve this concentration.
Rearranging to solve for the number of moles of solute,
Moles of solute = Molality * kg of solvent
Which you can now plug in your known values to solve for.

Once we know the number of moles of solute needed, the last step (converting to grams of solute) is easy…simply multiply by the molar mass of Sulfur and you will get the number of grams of Sulfur needed to lower the freezing point of the CCl4 by 2.10 degrees.