Calculation of hypochlorite concentration from active chlorine

So in the lab we got sodium hypochlorite solution. The concentration is in active chlorine (13%) on the label. How do I calculate it to concentration of hypochlorite? I need certain amount of hypochlorite (0,868g roughly) for the reaction I am trying to do and because it is not going as described in the article as they were using 5% hypochlorite solution I am checking for possible miscalculations on my part.


Thank you for the help and answers in advance




The active chlorine, in %, is the mass-percentage of $\ce{Cl2}$
dissolved in the solution, without taking into account the fact that this $\ce{Cl2}$ may be at least partially transformed into $\ce{ClO-}$ or $\ce{HClO}$, according to one of the two equations depending upon the pH value : $$\ce{Cl2 + 2 OH^- -> ClO^- + Cl- + H2O}$$ $$\ce{Cl2 + H2O -> HClO + Cl- + H+}$$


One liter solution with $13$% active chlorine has been made with $130$ g $\ce{Cl2}$ being dissolved into enough $\ce{NaOH}$ solution. And $130$ g $\ce{Cl2}$ is $\frac{130 g}{71 g/mol} = 1.83$ mol $\ce{Cl2}$. This solution has a concentration [$\ce{ClO^-}] = 1.83$ M.


The origin of the word "active" is coming from the fact that, if such a solution is acidified with some strong acid, the previous equations will be reversed, and $1$ L solution with $13$% active chlorine will liberate $130$ g chlorine $\ce{Cl2}$.


Pleas take notice that sodium hypochlorite cannot be obtained easily as a dry substance by evaporating its solution. If such a solution is evaporated by heating it, $\ce{NaClO}$ will disproportionate according to $$\ce{3 ClO^- -> ClO3^- + 2 Cl^- }$$


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