# Question #ca323

##### 1 Answer

#### 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*,

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

So, **percent concentration by mass**, *grams of solute* you get **per**

In your case, a **for every**

#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 **for every**

#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 thecorrect number of sig figs