# Question ca323

Apr 12, 2016

$\text{24% m/m}$

#### Explanation:

The idea here is that you need to use the percent concentration of the two solution to determine how much solute, which in your case is ammonium chloride, $\text{NH"_4"Cl}$, they contain.

Once you know that, use the mass of the resulting solution to find its percent concentration by mass.

So, percent concentration by mass, $\text{% m/m}$, essentially tells you how many grams of solute you get per $\text{100 g}$ of solution.

In your case, a $\text{5% m/m}$ ammonium chloride solution will contain $\text{5 g}$ of ammonium chloride for every $\text{100 g}$ of solution. This means that your sample will contain

200 color(red)(cancel(color(black)("g solution"))) * overbrace(("5 g NH"_4"Cl")/(100color(red)(cancel(color(black)("g solution")))))^(color(purple)("5% m/m")) = "10 g NH"_4"Cl"

Do the same for the second solution, which will contain $\text{30 g}$ of ammonium chloride for every $\text{100 g}$ of solution

600color(red)(cancel(color(black)("g solution"))) * overbrace(("30 g NH"_4"Cl")/(100color(red)(cancel(color(black)("g solution")))))^(color(purple)("30% m/m")) = "180 g NH"_4"Cl"

So, how much ammonium chloride will the target solution contain?

${m}_{N {H}_{4} C l} = \text{10 g" + "180 g" = "190 g NH"_4"Cl}$

The total mass of the solution will now be

m_"total" = m_(5%) + m_(30%)

${m}_{\text{total" = "200 g" + "600 g" = "800 g}}$

Therefore, the percent concentration by mass in this target solution will be

"% NH"_4"Cl" = (190 color(red)(cancel(color(black)("g"))))/(800color(red)(cancel(color(black)("g")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)24%color(white)(a/a)|)))

I'll leave the answer rounded to two sig figs, even though the values given to you only justify one sig fig for the answer, i.e.

"% NH"_4"Cl" = 20% -># rounded to the correct number of sig figs