Question

A certain liquid X has a normal boiling point of 108.30 °C and a boiling point elevation constant K =1.07 °C kg mol^-1. A solution is prepared by dissolving some ammonium sulfate ((NH4)2SO4) in 600. g of X. This solution boils at 109.7 °C. Calculate the mass of (NH4)2SO4 that was dissolved.

Answer 1

103.7 g

Step-by-step explanation:

We know that the change in the boiling temperature is given by

ΔTb = Kb x m where Kb is the molal boiling point elevation constant, and m is the molality of the solution.

The question gives us all the information required to calculate the molality of the solution. From the molality concentration we can find the number of moles, and hence the mass once we multiply the molecular weight.

ΔTb = 109.7 ºC - 108.30 ºC = 1.4 ºC

ΔTb = Kb x m ⇒ m = ΔTb / Kb

m = 1.4 ºC / 1.07 ºC kgmol⁻¹ = 1.3 mol /kg

molality is the number of moles divided into kg of solvent in this case solvent X, so

moles (NH₄)₂SO₄ that must have dissolved in 600 g (0.600 kg of solvent) are given by:

0.600 kg x 1.3 mol /kg = 0.79 mol

and the mass:

0.79 mol x 132.14 g/mol = 103.7 g