How to Calculate Molarity (With Dilution Examples)

$$M = \frac{\text{moles of solute}}{\text{liters of solution}}$$

The unit is mol/L, written as M and read as "molar." A 2.0 M solution has 2.0 moles of solute in every liter of solution.

Two details matter:

1. The volume is the volume of the whole solution, not just the water you added.
2. The volume must be in liters. Divide milliliters by 1000: 250.0 mL = 0.2500 L.

The formula can be rearranged to find any one of the three quantities:

$$\text{moles} = M \times V \qquad V = \frac{\text{moles}}{M}$$

Problem: 5.85 g of NaCl is dissolved in enough water to make 250.0 mL of solution. What is the molarity?

1. Find the molar mass: Na 22.99 + Cl 35.45 = 58.44 g/mol.
2. Convert grams to moles:

$$5.85 \text{ g} \div 58.44 \text{ g/mol} = 0.1001 \text{ mol NaCl}$$

3. Convert mL to L: $250.0 \text{ mL} \div 1000 = 0.2500$ L.
4. Divide:

$$M = \frac{0.1001 \text{ mol}}{0.2500 \text{ L}} = 0.400 \text{ M}$$

Answer: 0.400 M NaCl (three significant figures, set by 5.85 g).

Problem A: How many grams of NaOH do you need to make 500. mL of 0.150 M NaOH?

1. Volume in liters: 0.500 L.
2. Moles: $0.150 \text{ mol/L} \times 0.500 \text{ L} = 0.0750$ mol.
3. Grams: $0.0750 \text{ mol} \times 40.00 \text{ g/mol} = 3.00$ g of NaOH.

To make it, dissolve 3.00 g of NaOH in some water, then add water until the total volume reaches 500. mL.

Problem B: What volume of 2.00 M HCl contains 0.250 mol of HCl?

$$V = \frac{0.250 \text{ mol}}{2.00 \text{ mol/L}} = 0.125 \text{ L} = 125 \text{ mL}$$

A good habit: write units on every number. If you end up with $\text{L}^2$ or $\text{mol}^2$, a step is upside down.

When you add water to a solution, the moles of solute stay the same. Only the volume changes. Since moles = $M \times V$, the moles before equal the moles after:

$$M_1V_1 = M_2V_2$$

• $M_1$, $V_1$: concentration and volume of the stock (concentrated) solution.
• $M_2$, $V_2$: concentration and volume of the diluted solution.

Here the volumes can stay in mL, as long as both are in the same unit.

Example 1: How much 6.0 M HCl do you need to make 250 mL of 1.5 M HCl?

1. Write what you know: $M_1 = 6.0$ M, $M_2 = 1.5$ M, $V_2 = 250$ mL. Solve for $V_1$.
2. Rearrange: $V_1 = \frac{M_2V_2}{M_1}$.
3. Plug in: $V_1 = \frac{(1.5)(250)}{6.0} = 62.5$ mL, about 63 mL to two significant figures.
4. Measure 62.5 mL of the stock and add water until the total volume is 250 mL. You add about 187.5 mL of water, not 250 mL.

Example 2: 50.0 mL of 3.00 M NaCl is diluted to 200.0 mL. What is the new concentration?

$$M_2 = \frac{M_1V_1}{V_2} = \frac{(3.00)(50.0)}{200.0} = 0.750 \text{ M}$$

Sanity check: the volume went up 4 times, so the concentration should drop to one fourth. $3.00 \div 4 = 0.750$. It matches.

Ionic compounds split into ions when they dissolve, so the ion concentration can be higher than the compound's molarity.

$$\mathrm{CaCl_2 \rightarrow Ca^{2+} + 2Cl^-}$$

In 0.20 M $\mathrm{CaCl_2}$:

1. $[\mathrm{Ca^{2+}}] = 0.20$ M, because each formula unit gives one $\mathrm{Ca^{2+}}$.
2. $[\mathrm{Cl^-}] = 2 \times 0.20 = 0.40$ M, because each formula unit gives two $\mathrm{Cl^-}$.

Square brackets, like $[\mathrm{Cl^-}]$, are shorthand for "molar concentration of." Paste it into the VeilGPT solver to check your steps on a multi-part solution problem.