# 800. mL of solution has a concentration of 3.0 M. How much water needs to be added to dilute the solution to 1.9 M?

##### 1 Answer

#### Answer:

#### Explanation:

As you know, a **dilution** is used to **decrease** the concentration of a given solution.

In order to *dilute* a given solution, you must make sure that you keep the *number of moles of solute* present **constant** and that you **increase** the total volume of the solution, usually by adding more *solvent*.

Since **molarity** is calculated by taking the *number of moles of solute* present **per liter** of solution, increasing the volume while keeping the number of moles of solute constant will result in a **decrease** in concentration.

Mathematically, dilution calculations can be performed using the following equation

The equation for dilution calculations looks like this

#color(blue)(overbrace(c_1 xx V_1)^(color(red)("moles of solute in concentrated solution")) = overbrace(c_2 xx V_2)^(color(red)("moles of solute in diluted solution"))#

Here

Now, you can rearrange this equation to find

#c_1V_1 = c_2V_2 implies c_1/c_2 = V_2/V_1#

This gives you the solution's **dilution factor**, *diluted solution*

#color(blue)(|bar(ul(color(white)(a/a)"D.F." = c_1/c_2 = V_2/V_1color(white)(a/a)|)))#

In your case, the concentration of the solution decreases from

#"D.F." = (3.0 color(red)(cancel(color(black)("M"))))/(1.9color(red)(cancel(color(black)("M")))) = 1.58#

So, if the stock solution was **times more concentrated** than the diluted solution, it follows that the *volume* of the diluted solution **must have increased** by a factor of

#"D.F." = V_2/V_1 implies color(purple)(|bar(ul(color(white)(a/a)color(black)(V_2 = "D.F." xx V_1)color(white)(a/a)|)))#

Therefore, the total volume of the diluted solution must be equal to

#V_2 = 1.58 * "800. mL" = "1264 mL"#

The volume of water needed will thus be

#V_2 = V_1 + V_"water"#

#V_"water" = "1264 mL" - "800. mL" = "464 mL"#

I'll leave the answer rounded to three **sig figs**, despite the fact that you only have two sig figs for the concentration of the two solutions

#V_"water" = color(green)(|bar(ul(color(white)(a/a)"464 mL"color(white)(a/a)|)))#