Question

At what temperature is the concentration of a saturated solution of `KCl` (molar mass 74.5 g) approximately 3 molal?

Answer 1

`0^@"C"`

Explanation:

In order to be able to answer this question, you need to have the solubility graph of potassium chloride, `"KCl"`, which looks like this

![)

Since the solubility of potassium chloride is given per `"100 g"` of water, calculate how many moles of potassium chloride would make a `"3.5-molal"` solution in that much water.

Keep in mind that molality is defined as moles of solute, which in your case is potassium chloride, divided by kilograms** of solvent, which in your case is water.

`color(blue)(b = n_"solute"/m_"solvent")`

This means that you have

`n_"solute" = b * m_"solvent"`

`n_"solute" = "3.5 mol" color(red)(cancel(color(black)("kg"^(-1)))) * 100 * 10^(-3)color(red)(cancel(color(black)("kg"))) = "0.35 moles KCl"`

Next, use potassium chloride's molar mass to figure out how many grams would contain this many moles

`0.35color(red)(cancel(color(black)("moles KCl"))) * "74.5 g"/(1color(red)(cancel(color(black)("mole KCl")))) = "26.1 g"`

Now take a look at the solubility graph and decide at which temperature dissolving `"26.1 g"` of potassium chloride per `"100 g"` of water will result in a saturated solution.

Practically speaking, you're looking for the temperature at which the saturation line, drawn on the graph in `color(blue)("blue")`, matches the value `"26.1 g"`.

From the look of it, dissolving this much potassium chloride per `"100 g"` of water will produce a `"3.5-molal"` saturated solution at `0^@"C"`.