Calculate the concentration of anhydrous sodium carbonate ( extNa2 extCO3) given a mass of 1.25 g dissolved in 250 cm3 of solution. Atomic masses: Na=23, C=12, O=16.

The atomic masses are given as Na = 23, C = 12, O = 16.

$$M(\text{Na}_2\text{CO}_3) = (2 \times M(\text{Na})) + M(\text{C}) + (3 \times M(\text{O}))$$ $$M(\text{Na}_2\text{CO}_3) = (2 \times 23) + 12 + (3 \times 16)$$ $$M(\text{Na}_2\text{CO}_3) = 46 + 12 + 48$$ $$M(\text{Na}_2\text{CO}_3) = 106 \text{ g/mol}$$

We assume the mass of sodium carbonate weighed is 1.25 g.

$$\text{Moles} = \frac{\text{Mass}}{\text{Molar mass}}$$ $$\text{Moles of Na}_2\text{CO}_3 = \frac{1.25 \text{ g}}{106 \text{ g/mol}}$$ $$\text{Moles of Na}_2\text{CO}_3 \approx 0.011792 \text{ mol}$$

The solution is made up to 250 cm$^3$.

$$1 \text{ dm}^3 = 1000 \text{ cm}^3$$ $$\text{Volume} = \frac{250 \text{ cm}^3}{1000 \text{ cm}^3/\text{dm}^3}$$ $$\text{Volume} = 0.250 \text{ dm}^3$$

$$\text{Concentration} = \frac{\text{Moles}}{\text{Volume}}$$ $$\text{Concentration of Na}_2\text{CO}_3 = \frac{0.011792 \text{ mol}}{0.250 \text{ dm}^3}$$ $$\text{Concentration of Na}_2\text{CO}_3 \approx 0.047168 \text{ mol/dm}^3$$

Rounding to three significant figures:

$$\text{Concentration of Na}_2\text{CO}_3 \approx 0.0472 \text{ mol/dm}^3$$

The concentration in mol dm$^{-3}$ of the anhydrous sodium carbonate is approximately $\boxed{\text{0.0472 mol/dm}^3}$.