Question

Question #ca323

Answer 1

`"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, `"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, `"% m/m"`, essentially tells you how many grams of solute you get per `"100 g"` of solution.

In your case, a `"5% m/m"` ammonium chloride solution will contain `"5 g"` of ammonium chloride for every `"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 `"30 g"` of ammonium chloride for every `"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_(NH_4Cl) = "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_"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