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

##0^@”C”##

In order to be able to answer this question, you need to have the 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 is defined as moles of , which in your case is potassium chloride, divided by kilograms** of , 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”##.

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