We have that the final molarity of H2O2 if 5.5 mL of a 3.0% w/w H2O2 solution, which has a density of 1.0 g/mL, is added to 5.5 mL of a starch-iodide solution. is
[tex]M=0.441[/tex]
From the question we are told
Calculate the final molarity of H2O2 if 5.5 mL of a 3.0% w/w H2O2 solution, which has a density of 1.0 g/mL, is added to 5.5 mL of a starch-iodide solution.
Generally the equation for the 3\% mass is mathematically given as
Let 3\% mass H2O2 be Hm
Therefore
[tex]3\%=\frac{Solute mass}{5.5ml*1g/ml}*1000\\\\Solute mass =X[/tex]
[tex]X=0.165g[/tex]
Generally the equation for the Molarity is mathematically given as
[tex]M=\frac{moles of H_2O_2}{V}[/tex]
Where
[tex]moles of H_2O_2=\frac{0.165}{39}\\\\moles of H_2O_2=4.2*10^{-3}[/tex]
Therefore
[tex]M=\frac{4.2*10^{-3}}{5.5*5.5*10^{-3}}[/tex]
[tex]M=0.441[/tex]
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