Question

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

Answer 1

`"220 g/100 g H"_2"O"`

Explanation:

A compound's solubility in water is usually expressed in grams per `"100 g"` of water, or, alternatively, in grams per `"100 mL"` of water.

In your case, you can use the information given to you to find the solubility of silver nitrate, `"AgNO"_3`, in water at `20^@"C"`, in grams per `"100 g"` of water, `"g/100 g H"_2"O"`.

So, you know that you can dissolve `"11.1 g"` of silver nitrate in `"5.0 g"` of water at `20^@"C"`. You can use this as a conversion factor to determine how much silver nitrate can be dissolved in `"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 `"11.1 g"` of silver nitrate.