# Question 76da3

Jan 30, 2017

${\text{252 g KNO}}_{3}$

#### Explanation:

The first thing to do here is to figure out how much potassium nitrate you can dissolve in water at ${75}^{\circ} \text{C}$ in order to have $\text{550.0 g}$ of saturated solution.

You already know that at ${75}^{\circ} \text{C}$, a saturated solution of potassium nitrate holds $\text{155 g}$ of solute for every $\text{100 g}$ of water.

In other words, you get $\text{155 g}$ of potassium nitrate for every

$\text{155 g KNO"_3 + "100.0 g H"_2"O" = "255 g solution}$

You can now use this as a conversion factor to figure out how much potassium nitrate you get in $\text{550 g}$ of saturated solution

550.0 color(red)(cancel(color(black)("g solution"))) * "155 g KNO"_3/(255color(red)(cancel(color(black)("g solution")))) = "334.3 g KNO"_3

You can now say that at ${75}^{\circ} \text{C}$, $\text{550.0 g}$ of saturated solution contain $\text{334.3 g}$ of potassium nitrate and

$\text{550.0 g solution " - " 334.3 g KNO"_3 = "215.7 g H"_2"O}$

Now, when you cool the solution to ${25}^{\circ} \text{C}$, you decrease the solubility of the solute to $\text{38.0 g}$ for every $\text{100.0 g}$ of water.

You know that this solution contains $\text{215.7 g}$ of water, so use the known solubility of potassium nitrate at ${25}^{\circ} \text{C}$ to calculate how much solute you can get in a saturated solution at this temperature

215.7 color(red)(cancel(color(black)("g H"_2"O"))) * "38.0 g KNO"_3/(100.0color(red)(cancel(color(black)("g H"_2"O")))) = "81.97 g KNO"_3

This means that your solution can hold a maximum of $\text{81.97 g}$ of potassium nitrate at ${25}^{\circ} \text{C}$. The rest of the solute will crystallize out of solution.

${m}_{\text{crystallize" = "334.3 g" - "81.97 g}}$

${m}_{\text{crystallize" = "252.3 g KNO}} _ 3$

Rounded to three sig figs, the answer will be

m_"crustallize" = color(darkgreen)(ul(color(black)("252 g KNO"_3)))#

Therefore, you can say that $\text{550.0 g}$ of saturated potassium nitrate solution will hold $\text{334.3 g}$ of potassium nitrate at ${75}^{\circ} \text{C}$ and $\text{81.97 g}$ at ${25}^{\circ} \text{C}$.

The difference between these two values will crystallize out of solution as a result of the decrease in solubility associated with the decrease in temperature.