An aqueous solution of 3.47 M silver nitrate, AgNO_3, has a density of 1.47 g/mL. What is percent by mass of AgNO_3 in the solution?

Nov 16, 2015

40.1%

Explanation:

Here's your strategy for this problem - you need to pick a sample volume of this solution, use the given density to find its mass, then the number of moles of silver nitrate to get mass of silver nitrate it contains.

So, to make the calculations easier, let's take a $\text{1.00-L}$ sample of this solution. Use the given density to determine its mass

1.00color(red)(cancel(color(black)("L"))) * (1000color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "1.47 g"/(1color(red)(cancel(color(black)("mL")))) = "1470 g"

Now, this $\text{1.00-L}$ solution will contain

$c = \frac{n}{V} \implies n = c \cdot V$

$n = 3.47 \text{moles"/color(red)(cancel(color(black)("L"))) * 1.00color(red)(cancel(color(black)("L"))) = "3.47 moles}$

Use silver nitrate's molar mass to help you determine how many grams of silver nitrate would contain this many moles

3.47color(red)(cancel(color(black)("moles AgNO"_3))) * "169.87 g"/(1color(red)(cancel(color(black)("mole AgNO"_3)))) = "589.4 g AgNO"_3

Now, the solution's percent concentration by mass is defined as the mass of the solute, in your case silver nitrate, divided by the mass of the solution, and multiplied by $100$.

$\textcolor{b l u e}{\text{%w/w" = "mass of solute"/"mass of solution} \times 100}$

Plug in your values to get

"%w/w" = (589.4color(red)(cancel(color(black)("g"))))/(1470color(red)(cancel(color(black)("g")))) xx 100 = color(green)("40.1%")