Question #a373a
1 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
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
Since every
#(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