# Question 678e3

Feb 28, 2017

$\text{0.15 moles KBr}$

#### Explanation:

The thing to remember about a solution's molarity is that it tells you how many moles of solute are present for every $\text{1 L}$ of solution.

In this case, a $\text{1.5-M}$ potassium bromide solution will contain $1.5$ moles of potassium bromide, the solute, for every $\text{1 L}$ of solution.

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{1.5 M KBr = 1.5 moles KBr in every 1 L of KBr solution}}}}$

Now, you know that

1 color(red)(cancel(color(black)("L"))) * (10^3"mL")/(1color(red)(cancel(color(black)("L")))) = "1000 mL"#

This means that your goal here is to figure out how many moles of potassium bromide you get in $\frac{1}{10} \text{th}$ of a liter of $\text{1.5 M}$ potassium bromide solution, i.e. in $\text{100 mL}$ of solution.

To do that, you can use the molarity of the solution as a conversion factor

$100 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{mL solution"))) * overbrace("1.5 moles KBr"/(1000color(red)(cancel(color(black)("mL solution")))))^(color(blue)("= 1.5 M KBr solution")) = color(darkgreen)(ul(color(black)("0.15 moles KBr}}}}$

I'll leave the answer rounded to two sig figs, but don't forget that you only have one significant figure for the volume of the sample.