Asolution was prepared by dissolving 39.0g of kcl in 225 g of water.
calculate the mass percent of kcl in the solution.
calculate the mole fraction of the ionic species kcl in the solution.
calculate the molarity of kcl in the solution if the total volume of the solution is 239 ml.
calculate the molality of kcl in the solution.

-  14.7% m/m

- Mole fraction of K⁺ = 0.039  

 Mole fraction of Cl⁻ = 0.039

-  2.19M

- 2.32 molal

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

it forms an anion that has a larger radius. anions have a negative charge and are larger than the original atom.