\( 100 \mathrm{~cm}^{3} \) of 0.100 molf(mm \( { }^{-3} \mathrm{HCl} \) solution is added to \( 400 \mathrm{~cm}^{3} \) of \( 0.05 \mathrm{modm}^{-3} \mathrm{HCl} \) solution. What is the concentration in moltm of the resuestant 5 olution

1. Convert the volumes from \(\mathrm{cm^3}\) to \(\mathrm{dm^3}\): \[ 100\,\mathrm{cm^3} = 0.100\,\mathrm{dm^3} \quad \text{and} \quad 400\,\mathrm{cm^3} = 0.400\,\mathrm{dm^3}. \] 2. Calculate the number of moles of \(\mathrm{HCl}\) in each solution: For the first solution: \[ n_1 = 0.100\,\mathrm{mol\,dm^{-3}} \times 0.100\,\mathrm{dm^3} = 0.010\,\mathrm{mol}. \] For the second solution: \[ n_2 = 0.05\,\mathrm{mol\,dm^{-3}} \times 0.400\,\mathrm{dm^3} = 0.020\,\mathrm{mol}. \] 3. Calculate the total number of moles of \(\mathrm{HCl}\) after mixing: \[ n_{\text{total}} = n_1 + n_2 = 0.010\,\mathrm{mol} + 0.020\,\mathrm{mol} = 0.030\,\mathrm{mol}. \] 4. Find the total volume of the mixed solution: \[ V_{\text{total}} = 0.100\,\mathrm{dm^3} + 0.400\,\mathrm{dm^3} = 0.500\,\mathrm{dm^3}. \] 5. Determine the concentration of the resulting solution: \[ C = \frac{n_{\text{total}}}{V_{\text{total}}} = \frac{0.030\,\mathrm{mol}}{0.500\,\mathrm{dm^3}} = 0.060\,\mathrm{mol\,dm^{-3}}. \]