# 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?

##### 1 Answer

Here's what I got.

#### Explanation:

You can actually use different approaches to figure out the molarity of the new solution.

As you know, a solution's molarity tells you how many *moles of solute* you get **per liter** of 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 **unchanged**, since you only added water to the stock solution.

Therefore,

#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.