# Question 1f4d1

Sep 22, 2016

About $\text{34 g}$.

#### Explanation:

You can solve this problem by looking at the solubility graph for potassium nitrate, ${\text{KNO}}_{3}$. 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 $\text{67 g/100 g H"_2"O}$. This means that in order to have a saturated solution of potassium nitrate at ${40}^{\circ} \text{C}$, you must dissolve about $\text{67 g}$ of this salt in $\text{100 g}$ of water.

In other words, an aqueous solution of potassium nitrate will contain a maximum of $\text{67 g}$ of dissolved potassium nitrate for every $\text{100 g}$ of water at ${40}^{\circ} \text{C}$.

You can now use this solubility to figure out how many grams of potassium nitrate can be dissolved in $\text{50 g}$ 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.