# The concentration of a solution of ammonia, "NH"_3 is "1.5% m/v". What is the molar concentration of a solution produced by diluting "25.0 mL" of this solution with "250 mL" of water?

Jul 27, 2017

${\text{0.080 mol L}}^{- 1}$

#### Explanation:

Start by calculating the number of moles of ammonia present in the $\text{25.0-mL}$ sample of $\text{1.5% m/v}$ ammonia solution.

As you know, a $\text{1.5% m/v}$ solution contains $\text{1.5 g}$ of solute, which in your case is ammonia, for every $\text{100 mL}$ of solution.

This means that your sample will contain

25.0 color(red)(cancel(color(black)("mL solution"))) * "1.5 g NH"_3/(100color(red)(cancel(color(black)("mL solution")))) = "0.375 g NH"_3

Use the molar mass of ammonia to convert this to moles

0.375 color(red)(cancel(color(black)("g"))) * "1 mole NH"_3/(17.031color(red)(cancel(color(black)("g")))) = "0.02202 moles NH"_3

Now, you're diluting this sample by adding $\text{250 mL}$ of water. The total volume of the diluted solution will be

$\text{25.0 mL + 250 mL = 275 mL}$

Since you only added water, the number of moles of ammonia will remain unchanged, i.e. the diluted solution contains the same number of moles as the $\text{25.0-mL}$ sample.

As you know, the molarity of a solution tells you the number of moles of solute present for every $\text{1 L} = {10}^{3}$ $\text{mL}$ of solution.

In your case, the number of moles of ammonia present in ${10}^{3}$ $\text{mL}$ of this diluted solution will be equal to

10^3 color(red)(cancel(color(black)("mL solution"))) * overbrace("0.02202 moles H"_3/(275 color(red)(cancel(color(black)("mL solution")))))^(color(blue)("the known composition of the diluted solution")) = "0.080 moles NH"_3

So, if you have $0.080$ moles of ammonia in ${10}^{3} \textcolor{w h i t e}{.} \text{mL" = "1 L}$ of the diluted solution, you can say that the molarity of the solution is equal to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{molarity = 0.080 mol L}}^{- 1}}}}$

The answer is rounded to two sig figs.