What is the concentration in moldm-³ of a solution, which contains 2.50g of calcium hydroxide per 500cm³ (CA=40, 0=16, H=1) find the solution

The chemical formula for calcium hydroxide is $\text{Ca(OH)}_2$.
Given atomic masses: $\text{Ca}=40$, $\text{O}=16$, $\text{H}=1$.
Molar mass of $\text{Ca(OH)}_2 = \text{Ca} + 2 \times (\text{O} + \text{H})$

$$= 40 + 2 \times (16 + 1)$$ $$= 40 + 2 \times 17$$ $$= 40 + 34$$ $$= 74 \text{ g mol}^{-1}$$

Mass of $\text{Ca(OH)}_2 = 2.50 \text{ g}$
Number of moles ($n$) = $\frac{\text{Mass}}{\text{Molar mass}}$

$$n = \frac{2.50 \text{ g}}{74 \text{ g mol}^{-1}}$$ $$n \approx 0.03378378 \text{ mol}$$

Given volume = $500 \text{ cm}^3$
Since $1 \text{ dm}^3 = 1000 \text{ cm}^3$:
Volume ($V$) in $\text{dm}^3 = \frac{500 \text{ cm}^3}{1000 \text{ cm}^3 \text{ dm}^{-3}}$

$$V = 0.500 \text{ dm}^3$$

Concentration ($C$) = $\frac{\text{Number of moles}}{\text{Volume}}$

$$C = \frac{0.03378378 \text{ mol}}{0.500 \text{ dm}^3}$$ $$C \approx 0.06756756 \text{ mol dm}^{-3}$$

Rounding to three significant figures:

$$C \approx 0.0676 \text{ mol dm}^{-3}$$

The concentration of the solution is $\text{0.0676 mol dm}^{-3}$.