In my lab we found OUR average molarity to be 0.83334 M, but we used 3%...

In my lab we found OUR average molarity to be 0.83334 M, but we used 3% H2O2 (hydrogen peroxide), so I'm not sure how to do this or what numbers to even use.

7. Convert the average molarity of hydrogen peroxide into % (wt) of hydrogen peroxide (the density of the solution = 1.009g/ml).

8. Comparing the above calculated % of hydrogen peroxide with the known (labeled on the bottle) value, calculate the % experimental error.

Solutions

Expert Solution

Solution :-

Average molarity = 0.83334 M

Lets assume we have 1 L solution then moles of H2O2 present in the solution are calculated as

Moles of H2O2 = molarity * volume in liter

= 0.83334 mol per L * 1 L

= 0.83334 mol H2O2

Now lets calculate the mass of the H2O2

Mass of H2O2 =moles * molar mass

= 0.83334 mol * 34.0147 g per mol

= 28.3 g H2O2

Now lets calculate the mass of the solution using the volume and density

Mass of solution = volume * density

= (1.0 L * 1000 ml / 1 L ) *1.009 g per ml

= 1009 g

Q7 ) Now lets calculate the percent of the H2O2

% wt H2O2 = (mass of H2O2/mass of solution )*100%

= (28.3 g / 1009g)*100%

= 2.8 %

The calculated percent of the H2O2 is 2.8 %

The % of H2O2 from the bottle is 3.0 %

8) % error = (calculated % wt - % wrom bottle / % wt from bottle )*100%

= (2.8 % - 3.0 %/3.0%) *100 %

= 6.66 %

So the percent error = 6.66 %

queen_honey_blossom answered 6 months ago

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