# Question #07095

Apr 29, 2016

${\text{16.9 g NH}}_{3}$

#### Explanation:

A solution's percent concentration by mass, or $\text{% m/m}$, essentially tells you how many grams of solute you get for every $\text{100 g}$ of solution.

In your case, an ammonia solution that is $\text{12.5% m/m}$ ammonia, ${\text{NH}}_{3}$, will contain $\text{12.5 g}$ of ammonia, which is the solute, for every $\text{100 g}$ of solution.

This means that you can essentially use the solution's percent concentration as a conversion factor to help you go from grams of solute to grams of solution or vice versa.

In your case, $\text{135 g}$ of this ammonia solution will contain

$135 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{g solution"))) * overbrace("12.5 g NH"_3/(100color(red)(cancel(color(black)("g solution")))))^(color(purple)("= 12.5% m/m")) = color(green)(|bar(ul(color(white)(a/a)"16.9 g NH}}_{3} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to three sig figs.

Now, you can use the exact same approach to find the volume of fruit juice present in $\text{250 mL}$ of a $\text{15% v/v}$ solution.

This time, you're dealing with percent concentration by volume, or $\text{% v/v}$, which essentially tells you the volume of solute that you get for every $\text{100 mL}$ of solution.

Use this concentration as a conversion factor to get the answer. You should end up with $\text{37.5 mL}$.