# Question #bfe91

Apr 13, 2015

So, you need to prepare a certain volume of a solution that has a certain molarity.

Molarity expresses the number of moles of solute, which you have to determine, dissolved per liter of solution.

$C = \frac{n}{V}$

If you have 1 mole of a solute dissolved in 1 L of solution, you'll have a 1 M solution

$C = \text{1 mole"/"1 L" = "1 mol/"L = "1 M}$

Since you know the volume and the molarity, you can rearrange the equation to solve for the number of moles that would give you that respective molarity

$C = \frac{n}{V} \implies n = C \cdot V$

For the above example, if you are given 1 L of a 1-M solution, the number of moles of solute will be

$n = 1 \text{mol"/cancel("L") * 1cancel("L") = "1 mole}$ Keep in mind that you must always use liters for the volume; even if you are given the volume in mililiters, you must convert it to liters when applying the molarity equation.

Here's another example. Let's say I give you 100 mL of a 1-M solution and ask you to determine the number of moles of solute present.

You'd get

$n = C \cdot V = 1 \text{mol"/cancel("L") * 100 * 10^(-3)cancel("L") = "0.1 moles}$

Compare this with the 1-L, 1-M solution. The volume is now 10 times smaller, but the molarity is the same, which means that you must have 10 times fewer moles present.

If you add 1 mole to 100 mL, the molarity will be 10 times greater than when you had 1 mole in 1 L because you get the same amount of moles in a smaller volume $\to$ increased molarity.

$C = \frac{n}{V} = \text{1 mole"/(100 * 10^(-3)"L") = "10 mol/L" = "10 M}$