**Step 1. **total mol

NaCl → Na^{+} + Cl^{–}

$\mathbf{0}\mathbf{.}\mathbf{12}\mathbf{}\overline{)\mathbf{L}}\mathbf{\times}\frac{\mathbf{0}\mathbf{.}\mathbf{13}\mathbf{}\overline{)\mathbf{mol}\mathbf{}\mathbf{NaCl}}}{\overline{)\mathbf{L}}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{Cl}}^{\mathbf{-}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}\mathbf{NaCl}}}\mathbf{=}$**0.0156 mol Cl**^{–}

MgCl_{2} → Mg^{2+} + 2 Cl^{–}

A solution is prepared by mixing 0.12 L of 0.13 M sodium chloride with 0.24 L of a 0.18 M MgCl_{2} solution.

What volume of a 0.22 M silver nitrate solution do we need to precipitate all the Cl^{− }ion in the solution as AgCI?

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