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

Feb 10, 2016

${0}^{\circ} \text{C}$

#### Explanation:

In order to be able to answer this question, you need to have the solubility graph of potassium chloride, $\text{KCl}$, which looks like this Since the solubility of potassium chloride is given per $\text{100 g}$ of water, calculate how many moles of potassium chloride would make a $\text{3.5-molal}$ 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.

$\textcolor{b l u e}{b = {n}_{\text{solute"/m_"solvent}}}$

This means that you have

${n}_{\text{solute" = b * m_"solvent}}$

${n}_{\text{solute" = "3.5 mol" color(red)(cancel(color(black)("kg"^(-1)))) * 100 * 10^(-3)color(red)(cancel(color(black)("kg"))) = "0.35 moles KCl}}$

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 $\text{26.1 g}$ of potassium chloride per $\text{100 g}$ 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 $\textcolor{b l u e}{\text{blue}}$, matches the value $\text{26.1 g}$.

From the look of it, dissolving this much potassium chloride per $\text{100 g}$ of water will produce a $\text{3.5-molal}$ saturated solution at ${0}^{\circ} \text{C}$.