How would you prepare 250 mL of 0.158 M solution of #MgCl_2# from a 3.15 M stock solution?
Here's what I got.
The idea here is that you need to figure out how many moles of magnesium chloride,
#color(blue)(c = n/V)#
So, how many moles of magnesium chloride must be present in the target solution?
#c = n/V implies n = c * V#
#n = "0.158 M" * 250.0 * 10^(-3)"L" = "0.0395 moles MgCl"_2#
Now determine what volume of the target solution would contain this many moles of magnesium chloride
#c = n/V implies V = n/c#
#V = (0.0395color(red)(cancel(color(black)("moles"))))/(3.15color(red)(cancel(color(black)("moles")))/"L") = "0.01254 L"#
Rounded to three sig figs and expressed in mililiters, the volume will be
#V = "12.5 mL"#
So, to prepare your target solution, use a
This is equivalent to diluting the