\( 10.0 \mathrm{dm}^{3} \) of water was added to \( 2.0 \mathrm{~mol} \mathrm{dm}^{-3} \) of \( 2.5 \mathrm{dm}^{3} \) solution of HCl . What is the concentration of the final solution in \( \mathrm{mol} \mathrm{dm}^{-3} \) ?

1. **Calculate the moles of HCl originally present:** The moles of HCl are given by \[ \text{moles} = \text{concentration} \times \text{volume} \] Substituting the given values: \[ \text{moles} = 2.0\, \mathrm{mol\,dm^{-3}} \times 2.5\, \mathrm{dm^3} = 5.0\, \mathrm{mol} \] 2. **Determine the final volume of the solution:** The final volume is the sum of the initial solution volume and the volume of water added: \[ \text{final volume} = 2.5\, \mathrm{dm^3} + 10.0\, \mathrm{dm^3} = 12.5\, \mathrm{dm^3} \] 3. **Calculate the concentration of the final solution:** The concentration after dilution is calculated using \[ \text{concentration} = \frac{\text{moles of solute}}{\text{final volume}} \] Therefore, \[ \text{concentration} = \frac{5.0\, \mathrm{mol}}{12.5\, \mathrm{dm^3}} = 0.4\, \mathrm{mol\,dm^{-3}} \] 4. **Final Answer:** The concentration of the final solution is \[ \boxed{0.4\, \mathrm{mol\,dm^{-3}}} \]