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

answer: i have one answer

acids are substances that ionize (break off) in an aqueous solution to produce hydrogen (h+) ions

explanation:

hope this and sorry i could only come up with one answer!

answer : one can design an experiment using a liquid and an additive that can be added to the liquid and observe the chemical changes in the reaction species.

one would need a beaker which is filled with a known amount of the liquid in which thermometer is placed to measure the temperature changes; after addition of the additive to the liquid solution one can note down the temperature change and find whether the reaction is endothermic or exothermic in nature. by subtracting the initial temperature value with the final temperature one can find the temperature changes.

various combinations of liquid and additive can be used. liquid in this case can be a known quantity of water and additives can be epsom salt or without epsom salt the experiment can be conducted to notice the changes. water and ice, hydrogen peroxide and dry yeast, vinegar and baking soda,etc.