Question #3a4cd

1 Answer
Jan 6, 2017

Answer:

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.

https://saylordotorg.github.io/text_general-chemistry-principles-patterns-and-applications-v1.0/s17-02-solubility-and-molecular-struc.html

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"#.