# Question #53829

##### 1 Answer

#### Explanation:

The idea here is that when a solution undergoes a **dilution**, its final volume **increases** by a factor of

This is done by adding enough **solvent** to ensure that the total volume of the solution increases by a factor of **decrease** by a factor of

In your case, the starting solution has a **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

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 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.