Question

Question #53829

Answer 1

`"1.67% m/v"`

Explanation:

The idea here is that when a solution undergoes a `1:3` dilution, its final volume increases by a factor of `3`.

This is done by adding enough solvent to ensure that the total volume of the solution increases by a factor of `3`. As a result, the concentration of the solution will decrease by a factor of `3`.

In your case, the starting solution has a `"5% m/v"` concentration and a volume of `"250 mL"`. The total volume of the solution after the dilution will be

`V_"final" = 3 xx "250 mL" = "750 mL"`

The concentration of the solution after the dilution will be

`"% m/v" = "5%"/3 = color(darkgreen)(ul(color(black)(1.67%)))`

You can double-check this result by working with the mass of solute present in the solution. In the initial solution, you have

`250 color(red)(cancel(color(black)("mL solution"))) * overbrace("5 g NaCl"/(100color(red)(cancel(color(black)("mL solution")))))^(color(blue)("= 5% m/v NaCl")) = "12.5 g NaCl"`

Remember, the solution is diluted by adding solvent, so you know for a fact that the diluted solution will contain `"12.5 g"` of sodium chloride.

In order to find the diluted solution's mass by volume percent concentration, you must determine how many grams of solute you have in `"100 mL"` of solution. The diluted solution has a volume of `"750 mL"` and contains `"12.5 g"` of sodium chloride, which means that `"100 mL"` of this solution will contain

`100 color(red)(cancel(color(black)("mL solution"))) * "12.5 g NaCl"/(750 color(red)(cancel(color(black)("mL solution")))) = "1.67 g NaCl"`

This means that the mass by volume percent concentration of the diluted solution will be

`color(darkgreen)(ul(color(black)("% m/v = 1.67% NaCl")))`

I'll leave the answer rounded to three sig figs, but keep in mind that you only have one significant figure for the concentration of the initial solution.