$V_{Cl_2} = 220 \text{ mL} = 0.220 \text{ L}$.

    $n_{Cl_2} = \frac{V_{Cl_2}}{24.79}$

    $n_{Cl_2} = \frac{0.220}{24.79} \approx 0.0088745 \text{ mol}$

$V_{H_2O} = 100 \text{ mL} = 0.1 \text{ L}$.

    $V_{dd} \approx V_{H_2O} = 0.1 \text{ L}$

    $a = C_{M(Cl_2)} = \frac{n_{Cl_2}}{V_{dd}}$

    $a = \frac{0.0088745 \text{ mol}}{0.1 \text{ L}} \approx 0.088745 \text{ mol/L}$

    $a \approx 0.09 \text{ mol/L}$

`->`Vậy, giá trị của a là khoảng 0.09 mol/L.