Question #37461
1 Answer
See explanation.
Explanation:
For starters, it's worth mentioning that you won't find a volume by volume percent concentration,
A much more plausible percent concentration to use for a solution of sodium chloride is the mass by volume percent concentration,
#"% m/v" = "grams of solute"/"100 mL solution" xx 100%#
For, for example, if you have a
By comparison, the mass by mass percent concentration,
#"% m/m" = "grams of solute"/"100 g solution" xx 100%#
In this example, a
Notice that you can convert between the two percent concentrations by using the density of the solution. For example, if you have a solution of sodium chloride that has a density of
#100 color(red)(cancel(color(black)("mL solution"))) * (rho color(white)(.)"g solution")/(1color(red)(cancel(color(black)("mL solution")))) = (100 * rho)color(white)(.)"g"#
So, you know that
#"100 mL solution " -> " "(100 * rho) color(white)(.)"g solution"#
This means that you have
#"% m/v" = "grams of solute"/"100 mL solution" xx 100%#
#"% m/v" = "grams of solute"/((100 * rho)color(white)(.)"g solution") xx 100%#
This is equivalent to
#"% m/v" = overbrace("grams of solute"/"100 g solution" xx 100%)^(color(blue)("= %m/m")) * 1/rho#
which gets you
#color(blue)(ul(color(black)("% m/v" = 1/rho * "% m/m")))#
Notice that if the density of the solution is