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

## The solubility of $\text{KCl}$ is 34 g/100 g $\text{H"_2"O}$ at 20°C.

Aug 17, 2016

$\text{100 g}$

#### Explanation:

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

This means that at ${20}^{\circ} \text{C}$, a saturated solution of potassium chloride will contain $\text{34 g}$ of dissolved salt for every $\text{100 g}$ 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 $\text{34 g}$ of salt in $\text{100 g}$ of water at ${20}^{\circ} \text{C}$.

Now, your goal here is to figure out how much potassium chloride can be dissolved in $\text{300 g}$ 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

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{mass of KCl " = " 100 g}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$