# The solubility of KNO_3 is 155 g per 100 g of water at 75° Cand 38.0 g at 25°C. What mass (in grams) of KNO_3 will crystallize out of solution if exactly 100.0 g of its saturated solution at 75° C is cooled to 25°C?

Jan 4, 2016

$\text{45.9 g}$

#### Explanation:

So, you know that potassium nitrate, $\text{KNO} 3$, has a solubility of $\text{155 g}$ per $\text{100 g}$ of water at ${75}^{\circ} \text{C}$ and of $\text{38.0 g}$ at ${25}^{\circ} \text{C}$.

What that means is that at ${75}^{\circ} \text{C}$, you can dissolve as much as $\text{155 g}$ of solid potassium nitrate in water without the solution becoming saturated.

Once you hit that $\text{155 g}$ mark, the solution becomes saturated, which means that the solution can't dissolve any more solid.

Now, the same thing can be said for the solution at ${25}^{\circ} \text{C}$. At this temperature, dissolving less than $\text{38.0 g}$ of potassium nitrate per $\text{100 g}$ of water will result in an unsaturated solution.

At the $\text{38.0 g}$ mark, the solution will become saturated.

Now, take a look at how the solubility graph for potassium nitrate looks like

So, you are starting with $\text{100 g}$ of saturated solution at ${75}^{\circ} \text{C}$. Mind you, you have $\text{100 g}$ of solution that contains as much dissolved potassium nitrate as possible.

This solution will contain

100.0 color(red)(cancel(color(black)("g solution"))) * "155 g KNO"_3/((155 + 100)color(red)(cancel(color(black)("g solution")))) = "60.78 g KNO"_3

Now, potassium nitrate's solubility is given per $\text{100 g}$ of water. Calculate how much water you have in this initial solution

${m}_{\text{water" = "100.0 g" - "60.78 g" = "39.22 g}}$

Next, determine how much potassium nitrate can be dissolved in $\text{39.22 g}$ of water at ${26}^{\circ} \text{C}$ in order to make a saturated solution, i.e. have the maximum amount of dissolved potassium nitrate possible

39.22 color(red)(cancel(color(black)("g water"))) * "38.0 g KNO"_3/(100color(red)(cancel(color(black)("g water")))) = "14.9 g KNO"_3

This means that when the initial solution is cooled from ${75}^{\circ} \text{C}$ to ${25}^{\circ} \text{C}$, the amount of water that it contained will only hold $\text{14.9 g}$ of dissolved potassium nitrate.

The rest will crystallize out of solution

m_"crystallize" = "60.78 g" - "14.9 g" = color(green)("45.9 g")

The answer is rounded to three sig figs.