Question

Question #1f4d1

Answer 1

About `"34 g"`.

Explanation:

You can solve this problem by looking at the solubility graph for potassium nitrate, `"KNO"_3`.


For starters, the temperature is given to you in Kelvin, so convert it to degrees Celsius

`t[""^@"C"] = "313 K" - "273.15 K" ~~ 40^@"C"`

Now, according to the solubility graph, potassium nitrate has a solubility of about `"67 g/100 g H"_2"O"`. This means that in order to have a saturated solution of potassium nitrate at `40^@"C"`, you must dissolve about `"67 g"` of this salt in `"100 g"` of water.

In other words, an aqueous solution of potassium nitrate will contain a maximum of `"67 g"` of dissolved potassium nitrate for every `"100 g"` of water at `40^@"C"`.

You can now use this solubility to figure out how many grams of potassium nitrate can be dissolved in `"50 g"` of water at this temperature to have a saturated solution

`50 color(red)(cancel(color(black)("g H"_2"O"))) * "67 g KNO"_3/(100color(red)(cancel(color(black)("g H"_2"O")))) = "34 g KNO"_3`

I'll leave the answer rounded to two sig figs.