A sodium hydroxide solution is prepared by dissolving 80.0 g of sodium hydroxide (NaOH) pellets in 2 dm³ of water. What is the concentration of the solution?

The atomic masses are: Na = 23.0 g/mol, O = 16.0 g/mol, H = 1.0 g/mol.

$$\text{Molar mass of NaOH} = 23.0 + 16.0 + 1.0 = 40.0 \text{ g/mol}$$

$$\text{Moles} = \frac{\text{Mass}}{\text{Molar mass}}$$ $$\text{Moles of NaOH} = \frac{80.0 \text{ g}}{40.0 \text{ g/mol}} = 2.00 \text{ mol}$$

The volume of the solution is given as $2 \text{ dm}^3$.

$$\text{Concentration} = \frac{\text{Moles of solute}}{\text{Volume of solution}}$$ $$\text{Concentration} = \frac{2.00 \text{ mol}}{2 \text{ dm}^3} = 1.00 \text{ mol/dm}^3$$

The concentration of the solution is $1.00 \text{ mol/dm}^3$.