Question

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?

Answer 1

`"0.080 mol L"^(-1)`

Explanation:

Start by calculating the number of moles of ammonia present in the `"25.0-mL"` sample of `"1.5% m/v"` ammonia solution.

As you know, a `"1.5% m/v"` solution contains `"1.5 g"` of solute, which in your case is ammonia, for every `"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 `"250 mL"` of water. The total volume of the diluted solution will be

`"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 `"25.0-mL"` sample.

As you know, the molarity of a solution tells you the number of moles of solute present for every `"1 L" = 10^3` `"mL"` of solution.

In your case, the number of moles of ammonia present in `10^3` `"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 `10^3color(white)(.)"mL" = "1 L"` of the diluted solution, you can say that the molarity of the solution is equal to

`color(darkgreen)(ul(color(black)("molarity = 0.080 mol L"^(-1))))`

The answer is rounded to two sig figs.