An aqueous solution was prepared by dissolving 0.131g of a substance in 25.5g of water. The molality of the solution was determined by Freezing Point Depression to be 0.056m.

Question

What is the molecular weight of the solute?

Answer 1

The freezing point depression for determining molality is based on the amount of a solute in 1000 g of water. So, we have to scale the 0.131 g solute in 25.5 g H2O up to how many grams of solute that would be in 1000 g H2O.

0.131 g / 25.5 g H2O x 1000 g = 5.14 g solute.

We were given that the number of moles of solute per 1000 g H2O is 0.056. Now, we have enough information to calculate the molecular weight of the solute.

5.14 g solute/1000 g H2O / 0.056 moles/1000 g H2O = 91.8 g / mole (molecular weight).