Question

Question #1497c

Answer 1

At about `46^@"C"`.

Explanation:

In order to be able to answer this question, you must have some information about how the solubility of potassium nitrate, `"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 `"80 g"` of potassium nitrate will dissolve in `"100 mL"` of water. Notice that the solubility of the salt is given in grams per `"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 `"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 `color(darkgreen)("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^@"C"` and `50^@"C"`, a little closer to the `50^@"C"` mark.

You can thus say that you can dissolve `"80 g"` of potassium nitrate in `"100 g"` of water at about `46^@"C"`.