# Question 1497c

Feb 25, 2017

At about ${46}^{\circ} \text{C}$.

#### Explanation:

In order to be able to answer this question, you must have some information about how the solubility of potassium nitrate, ${\text{KNO}}_{3}$, changes with temperature.

A useful tool to have at your disposal is the solubility graph for potassium nitrate, which looks like this Your goal is to find the temperature at which $\text{80 g}$ of potassium nitrate will dissolve in $\text{100 mL}$ of water. Notice that the solubility of the salt is given in grams per $\text{100 g}$ of water, which means that you'll have to convert the volume of water to grams.

Since no additional information was provided, you can assume that the density of water is equal to ${\text{1.0 g mL}}^{- 1}$, which implies that your sample has a mass of

100 color(red)(cancel(color(black)("mL"))) * "1.0 g"/(1color(red)(cancel(color(black)("mL")))) = "100 g"#

Now, look for $80$ on the vertical axis of the graph and draw a horizontal line until you hit the solubility curve, which is shown here in $\textcolor{\mathrm{da} r k g r e e n}{\text{green}}$.

From the point of intersection, draw a vertical line until you intersect the horizontal axis, i.e. the axis that shows the temperature.

Your line should fall somewhere in between ${40}^{\circ} \text{C}$ and ${50}^{\circ} \text{C}$, a little closer to the ${50}^{\circ} \text{C}$ mark.

You can thus say that you can dissolve $\text{80 g}$ of potassium nitrate in $\text{100 g}$ of water at about ${46}^{\circ} \text{C}$.