A 1.430 g sample of washing soda, which is sodium carbonate decahydrate, Na₂CO₃· 10H₂O (286 g/mol)…

Step 1. $n(\text{HCl}) = 0.1000\ \text{mol/L} \times 0.00940\ \text{L} = 0.940\ \text{mmol}$.

Step 2. $\text{Na}_{2}\text{CO}_{3}\ +\ 2\text{HCl}\ \rightarrow \ 2\text{NaCl}\ +\ \text{H}_{2}\text{O}\ +\ \text{CO}_{2}$. So $n(\text{Na}_2\text{CO}_3) = \dfrac{0.940}{2} = 0.470\ \text{mmol}$ in the portion.

Step 3. The whole sample is ten times the portion: $4.70\ \text{mmol}$. The mass of the decahydrate is $4.70\ \text{mmol} \times 286\ \text{mg/mmol} = 1344\ \text{mg} = 1.344\ \text{g}$.

Step 4. $\%\ \text{purity} = \dfrac{1.344}{1.430} \times 100 = 94.0\ \%$.