When 12.5 g of K2SO4 dissolves in water, the actual species present will be the ions K+(aq) and SO42-
You cannot know both the molarity and molality without being given the density of the solution. The way the question is stated, it is the molarity that you can't find. If the solution is considered very dilute, then the two are the same and the volume of water can be taken as equal to the volume of solution.
Molality is defined as moles of solute per kg (essentially liter) of water.
Find moles of solute and liters of water and divide:
12.5 g K2SO4 (1mole/174.26 g) = .07173 moles of solute
Molality = .07173 moles/.050 liters water = 1.43 molal (This is NOT dilute)
The mass percent and mole percent of potassium sulfate are obtained from .07173 moles / (.07173 + 50 g/18.02 g/mole) x 100 and 12.5 g/(12.5 + 50g) x 100 respectively.
If we assume a density of 1.05 g/ml for the solution (about right from density/concentration of potassium sulfate plots)
The mass of the solution is 50 g water + 12.5g of potassium sulfate = 62.5 g
The volume of the solution is 62.5 g/(1.05 g/ml) = 59.52 ml
Molarity = .07173 moles/.05952 liters of solution = 1.21 Molar
The issue is that the presence of solute adds to the mass of the solution directly, but volume does not add the same way except for immiscible phases.
Hoped that helped. Take care.
