Question #ca323

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