Question

Question #3a4cd

Answer 1

Here's what I got.

Explanation:

Start by taking a look at the solubility graph for potassium nitrate, `"KNO"_3`

![www.mts.net]()

As you can see, the graph indicates the mass of potassium nitrate that can be dissolved per `"100 g H"_2"O"` at various temperatures.

The green curve shows the amount of potassium nitrate that can be dissolved in `"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.

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Now, start at `45^@"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^@"C"`, a saturated solution of potassium nitrate can hold about `"75 g"` of salt for every `"100 g"` of water. Any amount that exceeds this value will remain undissolved at this temperature.

In your case, you have `"95 g"` of potassium nitrate in `"100 g"` of water. You can say that you have an excess of

`overbrace(" 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 `"75 g"` of dissolved potassium nitrate, i.e. present as `"K"^(+)` and `"NO"_3^(-)` ions, and `"20 g"` of undissolved potassium nitrate, i.e. present as a solid, in `"100 g"` of water at `45^@"C"`.