Answer:
- 14.7% m/m
- Mole fraction of K⁺ = 0.039
Mole fraction of Cl⁻ = 0.039
- 2.19M
- 2.32 molal
Step-by-step explanation:
39.0g of KCl in 225 g of water
KCl is the solute
Water is the solvent
First of all let's find out the total mass of solution:
Solute mass + Solvent mass = Total mass of solution
39 g + 225 g = 264 g
In order to calculate the mass percent of KCl in the solution, we need to do this rule of three
264 g of solution have 39 g of solute
100 g of solution have (100 . 39)/264 = 14.7 g
As 39 grams are my mass of solute, I need the moles that this mass represents:
Mass / Molar mass = moles
39 g / 74.55 g/m = 0.523 moles
KCl → K⁺ + Cl⁻
To find out the mole fraction of the ionic species, we must know the dissociation of the salt (equation above), and the ratio is 1:1 so we have 0.523 moles of both, K⁺ and Cl⁻. We don't know the total moles, because we must consider moles that come from water.
Water mass / Water molar mass = Moles of water
225 g / 18 g/m = 12.5 moles
Total moles = Moles from water + Moles from salt
0.523 + 12.8 = 13.323
Mole fraction Cl⁻ = Moles Cl⁻ / Total moles
Mole fraction K⁺ = Moles K⁺ / Total moles
In both cases 0.523 / 13.323 == 0.039
Molarity is mol/L
We know the moles of solute, and we know the volume of solution (but this measure is in mL, we must convert to L)
239 mL = 0.239L
0.523 m/0.239L = 2.19M
Molality is mol/1kg of solvent
We know the moles of solute, and we know the mass of solvent, but this mass is in g; we should convert to kg
225 g = 0.225 kg
0.523 m/0.225kg = 2.32 molal