# What is the solubility of silver nitrate if only 11.1 g can dissolve in 5.0 g water at 20°C?

May 27, 2016

$\text{220 g/100 g H"_2"O}$

#### Explanation:

A compound's solubility in water is usually expressed in grams per $\text{100 g}$ of water, or, alternatively, in grams per $\text{100 mL}$ of water.

In your case, you can use the information given to you to find the solubility of silver nitrate, ${\text{AgNO}}_{3}$, in water at ${20}^{\circ} \text{C}$, in grams per $\text{100 g}$ of water, $\text{g/100 g H"_2"O}$.

So, you know that you can dissolve $\text{11.1 g}$ of silver nitrate in $\text{5.0 g}$ of water at ${20}^{\circ} \text{C}$. You can use this as a conversion factor to determine how much silver nitrate can be dissolved in $\text{100 g}$ of water at the same temperature

100color(red)(cancel(color(black)("g H"_2"O"))) * "11.1 g AgNO"_3/(5.0color(red)(cancel(color(black)("g H"_2"O")))) = "222 g"

This means that the solubility of the salt will be

"solubility AgNO"_3 = color(green)(|bar(ul(color(white)(a/a)color(black)("220 g/100 g H"_2"O")color(white)(a/a)|)))

The answer is rounded to two sig figs, the number of sig figs you have for the mass of water that can dissolve $\text{11.1 g}$ of silver nitrate.