Question

What is the maximum amount of `"KCl"` that can dissolve in 300 g of water at 20°C?

The solubility of `"KCl"` is 34 g/100 g `"H"_2"O"` at 20°C.

Answer 1

`"100 g"`

Explanation:

The problem provides you with the solubility of potassium chloride, `"KCl"`, in water at `20^@"C"`, which is said to be equal to `"34 g / 100 g H"_2"O"`.

This means that at `20^@"C"`, a saturated solution of potassium chloride will contain `"34 g"` of dissolved salt for every `"100 g"` of water.

As you know, a saturated solution is a solution that holds the maximum amount of dissolved salt. Adding more solid to a saturated solution will cause the solid to remain undissolved.

In your case, you can create a saturated solution of potassium chloride by dissolving `"34 g"` of salt in `"100 g"` of water at `20^@"C"`.

Now, your goal here is to figure out how much potassium chloride can be dissolved in `"300 g"` of water at this temperature. To do that, use the given solubility as a conversion factor to take you from grams of salt to grams of water

`300 color(red)(cancel(color(black)("g H"_2"O"))) * "34 g KCl"/(100color(red)(cancel(color(black)("g H"_2"O")))) = "102 g KCl"`

You should round this off to one sig fig, since that is how many sig figs you have for the mass of water

`color(green)(|bar(ul(color(white)(a/a)color(black)("mass of KCl " = " 100 g")color(white)(a/a)|)))`