# Question #a373a

##### 1 Answer

#### Answer:

#### Explanation:

We use **parts per million** to describe the concentration of a solution that contains **very small** amounts of solute, i.e. *trace amounts*.

The general idea behind using parts per million is that we look at the number of parts of solute present for every

#10^6 = 1,000,000#

parts of solution. In your case, a **parts** of solute **for every** **parts** of solution.

Now, you can assume that the density of this solution is equal to the density of water, which you can *approximate* as

#"0.1 g AgNO"_3 -> "0.1 parts of solute"#

for every

#10^6color(white)(.)"g solution"#

The mass of the solution will be equal to

#300 color(red)(cancel(color(black)("mL"))) * "1 g"/(1color(red)(cancel(color(black)("mL")))) = "300 g"#

At this point, you can use the solution's ppm concentration as a conversion factor to determine how many grams of silver(I) cations it must contain

#300 color(red)(cancel(color(black)("g solution"))) * overbrace("0.1 g Ag"^(+)/(10^6color(red)(cancel(color(black)("g solution")))))^(color(blue)("= 0.1 ppm solution")) = 3 * 10^(-5)# #"g Ag"^(+)#

This is equivalent to

#3 * 10^(-5) color(red)(cancel(color(black)("g"))) * (10^3 color(white)(.)"mg")/(1color(red)(cancel(color(black)("g")))) = "0.03 mg Ag"^(+)#

Now, in order to figure out how much *silver nitrate* will produce this much silver(I) cations in solution, use the compound's **percent composition**.

Silver nitrate has a **molar mass** of **molar mass** of

Since **every** **mole** of silver nitrate contains **mole** of silver, you can say that the percent composition of silver in this compound is

#(107.87 color(red)(cancel(color(black)("g mol"^(-1)))))/(169.87color(red)(cancel(color(black)("g mol"^(-1))))) * 100% = "63.5% Ag"#

This means that **every**

You can say that the mass of the silver(I) cations is approximately equal to the mass of elemental silver, which means that in order to get

#0.03 color(red)(cancel(color(black)("mg Ag"^(+)))) * "100 mg AgNO"_3/(63.5 color(red)(cancel(color(black)("mg Ag"^(+))))) = color(darkgreen)(ul(color(black)("0.05 mg AgNO"_3)))#

The answer is rounded to one **significant figure**.

Therefore, you can say that a solution that contains