# How many grams of solute are present in 795 mL of 0.870 M "KBr"?

Jul 28, 2016

$\text{82.3 g}$

#### Explanation:

The first thing to do here is use the molarity and volume of the solution to determine the number of moles of solute, which in your case is potassium bromide, $\text{KBr}$, it contains.

Once you know that, you can use the compound's molar mass to convert to grams.

So, you know that a $\text{0.870 M}$ potassium bromide solution will contain $0.870$ moles of potassium bromide for every liter of solution.

In your case, the solution is said to have a volume of

795 color(red)(cancel(color(black)("mL"))) * "1 L"/(10^3color(red)(cancel(color(black)("mL")))) = "0.795 L"

which means that it will contain

0.795 color(red)(cancel(color(black)("L solution"))) * "0.870 moles KBr"/(1color(red)(cancel(color(black)("L solution")))) = "0.69165 moles KBr"

Now, potassium bromide has a molar mass of ${\text{119.002 g mol}}^{- 1}$, which means that every mole of potassium bromide has a mass of $\text{119.002 g}$.

Your solution will thus contain

0.69165 color(red)(cancel(color(black)("moles KBr"))) * "119.002 g"/(1color(red)(cancel(color(black)("mole KBr")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("82.3 g")color(white)(a/a)|)))

The answer is rounded to three sig figs.