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QUESTION

# 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".