Question #1f3a2

The thing to remember about a solution's concentration in parts per million, or ppm, is that it tells you the number of grams of solute present for every

#10^6 = 1,000,000#

grams of the solution. In this case, the mass of silver(I) cations is so small compared to the mass of water that you can safely assume that the mass of the solution is equal to the mass of water.

So, you know that

#3 color(red)(cancel(color(black)("kg"))) * (10^3color(white)(.)"g")/(1color(red)(cancel(color(black)("kg")))) = 3 * 10^3# #"g"#

of this solution contain

#4.8 color(red)(cancel(color(black)("mg"))) * "1 g"/(10^3 color(red)(cancel(color(black)("mg")))) = 4.8 * 10^(-3)# #"g"#

of silver(I) cations. This means that #10^6# #"g"# of this solution will contain

#10^6 color(red)(cancel(color(black)("g solution"))) * (4.8 * 10^(-3)color(white)(.)"g Ag"^(+))/(3 * 10^3color(red)(cancel(color(black)("g solution")))) = "1.6 g Ag"^(+)#

Since this represents the number of grams of silver(I) cations present in #10^6# #"g"# of the solution, you can say that you have

#color(darkgreen)(ul(color(black)("concentration = 2 ppm Ag"^(+))))#

The answer must be rounded to one significant figure, the number of sig figs you have for the mass of water.