How to determine the equivalent weight of CO2
This is the danger of using formulae from books without fully understanding where the formulae came from.



So let's take it step by step. 

We have a titration of an unknown concentration solution with a known (standardised) solution.

The reaction $$\ce{2 NaOH + CO2 -&gt; Na2CO3 + H2O}$$ gives us the stoichiometric relationship between $\ce{NaOH}$ and $\ce{CO2}$, i.e. 2 moles of $\ce{NaOH}$ react with 1 mole of $\ce{CO2}$.

How much $\ce{CO2}$ Was used? Half the amount of $\ce{NaOH}$!

How much $\ce{NaOH}$ was used? 

$$\begin{align}
c(\ce{NaOH})  &amp;= \frac{\text{amount of substance}}{volume}\\
              &amp;= \frac{n_{\ce{NaOH}}}{V}\\
n_{\ce{NaOH}} &amp;= c(\ce{NaOH})\times V\\
n_{\ce{CO2}}  &amp;= \frac{1}{2} n_{\ce{NaOH}} \\
              &amp;= \frac{1}{2} c(\ce{NaOH})\times V\\
\end{align}$$

What is the concentration of $\ce{CO2}$ in sample?

$$c(\ce{CO2}) = \frac{n_{\ce{CO2}}}{\text{Volume of sample}} = \frac{1}{2}\frac{c(\ce{NaOH})\times V}{\text{Volume of sample}}$$

Concentration is also expressed as $\mathrm{mg/L}$. 1 mole of $\ce{CO2}$ weighs $44~\mathrm{g}$ or $44\times 1000$ mg so sticking this into the above

$$c(\ce{CO2})\mathrm{(mg/L)} = \frac{1}{2}\frac{c(\ce{NaOH})\times V}{\text{Volume of sample}} \times 44 \times 1000 = \frac{c(\ce{NaOH})\times V\times 22\times 1000}{\text{Volume of sample}}$$

That's where the $22$ came from.
This may not address exactly where the equivalent weight of $\ce{CO2}$ comes from but at least it shows you the derivation of how the $22$ got into the formula.