commercial bleach contains 5.25% sodium hypochlorite (NaOCl) by mass

density = 1.08 g/mL

assume 100 g solution:

mass NaOCl = 5.25 g

molar mass NaOCl = 74.44 g

Calculate moles of the hypochlorite ion, OCl^{-}:

NaOCl → Na^{+} + OCl^{-}

$\mathbf{moles}\mathbf{}\mathbf{NaOCl}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{25}\mathbf{}\overline{)\mathbf{g}\mathbf{}\mathbf{NaOCl}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\mathbf{mol}\mathbf{}\overline{)\mathbf{NaOCl}}}{\mathbf{74}\mathbf{.}\mathbf{44}\mathbf{}\overline{)\mathbf{g}\mathbf{}\mathbf{NaOCl}}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{OCl}}^{\mathbf{-}}}{\mathbf{1}\mathbf{}\mathbf{mol}\mathbf{}\overline{)\mathbf{NaOCl}}}$

**moles OCl ^{-} = 0.07052 mol**

Calculate volume (in L) of solution:

$\mathbf{volume}\mathbf{=}\mathbf{100}\mathbf{}\overline{)\mathbf{g}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\overline{)\mathbf{mL}}}{\mathbf{1}\mathbf{.}\mathbf{08}\mathbf{}\overline{)\mathbf{g}}}\mathbf{\times}\frac{{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\mathbf{L}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mL}}}$

**volume = 0.0926 L**

Calculate molarity of the solution:

Calculate the molarity of hypochlorite ion when 1.00 mL of commercial bleach is added to 25.00 mL of dye solution.

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