\( 10.0 \mathrm{dm}^{3} \) 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 \( \mathrm{mol} \mathrm{dm}^{-3} \) ?

To find the concentration of the final solution, we first determine the moles of HCl in the original solution. The initial solution has a concentration of \( 2.0 \, \text{mol} \, \text{dm}^{-3} \) and a volume of \( 2.5 \, \text{dm}^{3} \). Moles of HCl = Concentration × Volume = \( 2.0 \, \text{mol} \, \text{dm}^{-3} \times 2.5 \, \text{dm}^{3} = 5.0 \, \text{mol} \). Next, we calculate the total volume of the final solution after adding \( 10.0 \, \text{dm}^{3} \) of water: Total volume = \( 2.5 \, \text{dm}^{3} + 10.0 \, \text{dm}^{3} = 12.5 \, \text{dm}^{3} \). Now, we find the new concentration of HCl in the final solution: Concentration = Total moles / Total volume = \( 5.0 \, \text{mol} / 12.5 \, \text{dm}^{3} = 0.4 \, \text{mol} \, \text{dm}^{-3} \). So, the concentration of the final solution is **\( 0.4 \, \text{mol} \, \text{dm}^{-3} \)**.