What volume of 0.10 M #NaOH# can be prepared from 250 mL of 0.30 M #NaOH#?

The thing to notice here is that the soluzione iniziale is #3# times more concentrated rispetto target solution, da

#(0.30 color(red)(cancel(color(black)("M"))))/(0.10color(red)(cancel(color(black)("M")))) = 3#

Questo ti dice che il volume of the target solution must be #3# times greater than the volume of the initial solution.

That is the case because when you dilute a solution, you diminuire la sua concentrazione di crescente its volume while keeping the number of moles of soluto constant.

This implies that when the concentration of a solution diminuisce by a factor, which is usually called fattore di diluizione, #"DF"#, upon dilution, the volume of the diluted solution aumenta by the same factor #"DF"#.

#color(blue)(|bar(ul(color(white)(a/a)"DF" = V_"final"/V_"initial" = c_"initial"/c_"final"color(white)(a/a)|)))#

Quindi hai

#V_"final" = 3 xx "250 mL" = color(green)(|bar(ul(color(white)(a/a)color(black)("750 mL")color(white)(a/a)|)))#

La risposta è arrotondata a due sig fichi.

So, you can decrease the concentration of #"250 mL"# of sodium hydroxide solution from #"0.30 M"# a #"0.10 M"# by adding enough water to get its final volume to #"750 mL"#.