Chemistry · General chemistry I · Worked example

Find the molarity of a solution

11.7 g of NaCl is dissolved in water to make 500.0 mL of solution. What is the molarity of the solution?

11.7 g NaCl in 500.0 mL

Convert the solute to moles

The molar mass of NaCl is 22.99 + 35.45 = 58.44 g/mol, so 11.7 g is 11.7 ÷ 58.44 = 0.200 mol.

11.7 g NaCl \times \frac{1 mol NaCl}{58.44 g NaCl} = 0.200 mol NaCl

\begin{array}{c} 11.7 g NaCl \times \frac{1 mol NaCl}{58.44 g NaCl} \\ = 0.200 mol NaCl \end{array}

Molarity uses liters of solution: 500.0 mL is 0.5000 L.

500.0 mL = 0.5000 L

0.200 mol ÷ 0.5000 L = 0.400 mol/L, written 0.400 M. The answer has three significant figures, like 11.7 g.

M = \frac{0.200 mol}{0.5000 L} = 0.400 M

About a fifth of a mole in half a liter is about two fifths of a mole per liter, or 0.4 M. The answer fits.

The solution is 0.400 M NaCl.

How many grams of NaCl do you need to make 250.0 mL of 0.150 M NaCl?

Show the answer and explanation

Moles needed: 0.2500 L × 0.150 mol/L = 0.0375 mol. Mass: 0.0375 mol × 58.44 g/mol = 2.19 g.

0.2500 L \times \frac{0.150 mol}{1 L} \times \frac{58.44 g}{1 mol} = 2.19 g

\begin{array}{c} 0.2500 L \times \frac{0.150 mol}{1 L} \\ \times \frac{58.44 g}{1 mol} = 2.19 g \end{array}

Open the solution in Solutions & concentration and change the volume to 250.0 mL: the same salt in half the volume doubles the molarity, to 0.801 M.

Sources and scope

Authored study material. Tool results depend on the stated inputs and model assumptions.

Make it concrete

Open the example inputs, change a value and keep a useful result on your board.

Your existing work stays on this device. Examples open as editable copies.