Question #8ff57

Concentrations in parts per million, or ppm, are used when dealing with very, very small amounts of solutes.

More specifically, a concentration of #"1 ppm"# is equivalent to having a solution that contains one part solute for every #10^6# parts solvent.

When dealing with concentrations in ppm, it is safe to assume that the density of the solution will be equal to that of the solvent. In this case, if you take water's density to be equal to approximately #"1.00 g/mL"#, you can say that the mass of the sample will be equal to

#100color(red)(cancel(color(black)("mL"))) * "1 g"/(1color(red)(cancel(color(black)("mL")))) = "100 g"#

Mathematically, you can describe the concentration in ppm like this

#color(blue)("% ppm" = "mass of solute in grams"/"mass of solvent in grams" xx 10^6)#

Now, you need the solution to be #"20 ppm"# in sodium cations, #"Na"^(+)#. Calculate what mass of sodium would be needed to get to that concentration

#m_(Na^(+)) = (20 * m_"water")/10^6#

#m_(Na^(+)) = (20 * "100 g")/10^6 = 2000 * 10^(-6)"g" = 2 * 10^(-3)"g" = "2 mg"#

So, you need the solution to contain #"2 mg"# of sodium cations. Use sodium and sodium chloride's molar mass to help you find sodium chloride's percent composition

#( 22.989770 color(red)(cancel(color(black)("g/mol"))))/(58.44277color(red)(cancel(color(black)("g/mol")))) xx 100 = "39.34% Na"#

This means that very #"100 g"# of sodium chloride will contain #"39.34 g"# of sodium. In your case, you will get #"2 mg"# of sodium from

#2 color(red)(cancel(color(black)("mg Na"))) * "100 mg NaCl"/(39.34 color(red)(cancel(color(black)("mg Na")))) = "5.08 mg NaCl"#

Rounded to one sig fig, the answer will be

#m_"NaCl" = color(green)("5 mg")#

So, to get a solution that is #"20 pmm"# in sodium cations, you need to dissolve #"5 mg"# of sodium chloride in #"100 mL"# of water.