# Question 1a2c9

Jun 11, 2017

Here's what I got.

#### Explanation:

I'll show you how to find the solubility of potassium chloride and lead(II) nitrate, and leave the solubility of potassium dichromate to you as practice.

The idea here is that the solubility graph of a given salt tells you the amount of salt that can be dissolved in a specific volume of solvent and at a specific temperature in order to get a saturated solution.

In your case, I am assuming that the solubility graph expresses the solubility of three salts in grams per liter of water, ${\text{g L}}^{- 1}$, at temperatures that vary between $0$ and ${100}^{\circ} \text{C}$.

So, start by finding the $\text{40"^@"C}$ mark on the Temperature axis. For potassium chloride, go up from this point until you reach the $\textcolor{g r e e n}{\text{green}}$ line that shows the solubility of this salt at ${40}^{\circ} \text{C}$.

You should find that it lies at a little over ${\text{100 g L}}^{- 1}$, which means that at ${40}^{\circ} \text{C}$, you can dissolve a little over $\text{100 g}$ of potassium chloride for every $\text{1 L}$ of water in order to make a saturated solution of potassium chloride.

Since you're trying to dissolve $\text{500 g}$ of this salt, you will end up with a little over

$\text{500 g " - " 100 g" = "400 g}$

of undissolved salt. One significant figure will suffice here, so the answer will be $\text{400 g}$.

Do the same for lead(II) nitrate. Start at the ${40}^{\circ} \text{C}$ mark and move up until you reach the $\textcolor{b l u e}{\text{blue}}$ line that shows the solubility of this salt at ${40}^{\circ} \text{C}$.

The way I see it, the value lies right between $\text{700 g}$ and $\text{800 g}$ for every $\text{1 L}$ of water, so $\text{750 g}$ seems like a reasonable estimate.

Since the solution can hold a lot more lead(II) nitrate than you're trying to dissolve, i.e. $\text{750 g}$ vs $\text{500 g}$, you can say that all the mass of lead(II) nitrate will dissolve in this solution.

In other words, you will be left with $\text{0 g}$ of undissolved lead(II) nitrate.

Jun 11, 2017

Here's what I get.

#### Explanation:

For each salt, we find 40 °C on the x-axis and go straight up until we hit the graph for that salt.

bb("Pb"("NO"_3)_2)

The solubility of "Pb"("NO"_3)_2 at 40 °C is 750 g/L.

If we add 500 g of "Pb"("NO"_3)_2, it will all dissolve.

The amount of solid "Pb"("NO"_3)_2# remaining will be 0 g.

$\boldsymbol{\text{KCl}}$

The solubility of $\text{KCl}$ at 40 °C is 400 g/L.

If we add 500 g of $\text{KCl}$, only 400 g will dissolve.

The amount of solid $\text{KCl}$ remaining will be 100 g.

$\boldsymbol{{\text{K"_2"Cr"_2"O}}_{7}}$

The solubility of ${\text{K"_2"Cr"_2"O}}_{7}$ at 40 °C is 200 g/L.

If we add 500 g of ${\text{K"_2"Cr"_2"O}}_{7}$, only 200 g will dissolve.

The amount of solid ${\text{K"_2"Cr"_2"O}}_{7}$ remaining will be 300 g.