# Question #3a4cd

Jan 6, 2017

Here's what I got.

#### Explanation:

Start by taking a look at the solubility graph for potassium nitrate, ${\text{KNO}}_{3}$ As you can see, the graph indicates the mass of potassium nitrate that can be dissolved per $\text{100 g H"_2"O}$ at various temperatures.

The green curve shows the amount of potassium nitrate that can be dissolved in $\text{100 g}$ of water at a given temperature in order to produce a saturated solution, which as you know is a solution which holds the maximum amount of dissolved salt.

In other words, in a saturated solution, the rate at which the solid dissolves to produce ions in solution is equal to the rate at which the dissolved ions combine to reform the solid. Now, start at ${45}^{\circ} \text{C}$ and trace a vertical line up to the green curve. From the point of intersection, trace a horizontal line to the solubility value.

You should find that at ${45}^{\circ} \text{C}$, a saturated solution of potassium nitrate can hold about $\text{75 g}$ of salt for every $\text{100 g}$ of water. Any amount that exceeds this value will remain undissolved at this temperature.

In your case, you have $\text{95 g}$ of potassium nitrate in $\text{100 g}$ of water. You can say that you have an excess of

$\overbrace{\text{ 95 g ")^(color(blue)("what you want to dissolve")) - overbrace(" 75 g ")^(color(blue)("what can be dissolved")) = overbrace(" 20 g ")^color(blue)("what remains undissolved}}$

Therefore, your solution will contain $\text{75 g}$ of dissolved potassium nitrate, i.e. present as ${\text{K}}^{+}$ and ${\text{NO}}_{3}^{-}$ ions, and $\text{20 g}$ of undissolved potassium nitrate, i.e. present as a solid, in $\text{100 g}$ of water at ${45}^{\circ} \text{C}$.