If you dilute a stock solution that has a molarity of #"14.8 M"# and a volume of #"25 mL"# by adding #"100 mL"# of water, what will be the molarity of the resulting solution?
Here's what I got.
You can actually use different approaches to figure out the molarity of the new solution.
#color(blue)("molarity" = "moles of solute"/"liter of solution")#
The important thing to notice here is that you can decrease the concentration of a solution by keeping the number of moles of solute constant and increasing the volume of the solution.
This is exactly what is happening here. The volume of the solution will go from
#V_"total" = "25 mL" + "100 mL" = "125 mL"#
The volume of the solution increases by a factor of
#(125 color(red)(cancel(color(black)("mL"))))/(25color(red)(cancel(color(black)("mL")))) = 5#
which means that the molarity of the resulting solution will decrease by a factor of
#c_2 = c_1/5 = "14.8 M"/5 = color(green)("2.96 M")#
Essentially, you have diluted the stock solution by a factor of
As you know, diluting a solution implies keeping the number of moles of solute constant while increasing the total volume of the solution.
You actually have an equation to help you with dilution calculations
#color(blue)(overbrace(c_1V_1)^(color(purple)("moles of solute in tock solution")) = overbrace(c_2V_2)^(color(orange)("moles of solute in target solution")))" "#, where
Once again, the molarity of the diluted solution will come out to be
#c_1V_1 = c_2V_2 implies c_2 = V_1/V_2 * c_1#
#c_2 = (25 color(red)(cancel(color(black)("mL"))))/((25 + 100)color(red)(cancel(color(black)("mL")))) * "14.8 M" = color(green)("2.96 M")#
I'll leave the answer rounded to three sig figs, despite the fact that your values only justify one sig fig.