# How do you calculate the molarity of a solution prepared by dissolving 15.6 g of solid KBr in enough water to make 1.25 L of solution?

Jun 17, 2017

${\text{0.105 mol L}}^{- 1}$

#### Explanation:

For starters, convert the mass of potassium bromide to moles by using the molar mass of the compound.

15.6 color(red)(cancel(color(black)("g"))) * "1 mole KBr"/(119.002color(red)(cancel(color(black)("g")))) = "0.1311 moles KBr"

Now, in order to find the molarity of the solution, you need to figure out how many moles of solute are present for every $\text{1 L}$ of solution.

You already know that $\text{1.25 L}$ of solution contain $0.1311$ moles of potassium bromide, the solute, so all you have to do now is to use this known composition as a conversion factor to go from $\text{1 L}$ of solution to number of moles of solute.

1 color(red)(cancel(color(black)("L solution"))) * "0.1311 moles"/(1.25color(red)(cancel(color(black)("L solution")))) = "0.10488 moles KBr"

Since this represents the number of moles of solute present in $\text{1 L}$ of solution, you can say that the molarity of the solution is equal to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{molarity = 0.105 mol L}}^{- 1}}}}$

The answer must be rounded to three sig figs, the number of sig figs you have for your values.