# Calculate the amount of water, in grams, that must be added to prepare 16% by mass solution with "5.0 g" of urea, "(NH"_2)_2"CO" ?

Dec 3, 2017

$\text{26 g H"_2"O}$

#### Explanation:

The trick here is to realize that the solution's percent concentration by mass tells you the mass of solute present for every $\text{100 g}$ of the solution.

As you know, the mass of the solution includes the mass of the solute. In other words, the mass of the solution is given by the mass of the solute and the mass of the solvent.

So, a 16% by mass urea solution will contain $\text{16 g}$ of urea, the solute, for every $\text{100 g}$ of the solution. This means that the mass of the solution that contains $\text{5.0 g}$ of urea will be equal to

5.0 color(red)(cancel(color(black)("g urea"))) * overbrace("100 g solution"/(16color(red)(cancel(color(black)("g urea")))))^(color(blue)("= 16% by mass urea")) = "31.3 g solution"

Since you know that

overbrace("31.3 g")^(color(blue)("mass of solution")) = overbrace("5.0 g")^(color(blue)("mass of solute")) + "mass of solvent"

you can say that this solution will require

$\text{mass of solvent" = "31.3 g " - " 5.0 g}$

$\text{mass of solvent = 26.3 g}$

of water, your solvent. Rounded to two sig figs, the answer will be

"mass of water" = color(darkgreen)(ul(color(black)("26 g")))