Calculating Mass Lost in Decomposition of Hydrogen Peroxide

[nesan]

Someone check my answer please. :)

Homework Statement

The rate of decomposition of hydrogen peroxide was studied at a particular temperature. H2O2(aq) = H2O (l) + 1/2O2(g)

The initial concentration of hydrogen
peroxide was 0.200 mol/L. 10.0 s later,
it was measured to be 0.196 mol/L.

b) 0.500 L of hydrogen peroxide solution was
used for the experiment. What mass was lost
as O2 bubbled out of solution in this initial
10.0 s interval?

The Attempt at a Solution

What I did was.

0.500 L * 0.200 mol/L = 0.1

0.500 L * 0.196 mol/L = 0.098

0.1 - 0.098 = 0.002mol

0.002 * 16 = 0.032 g

What I'm doubting is.

0.500 L * 0.196 mol/L = 0.098

Won't the volume change after 10 seconds?

 

[Borek]

Yes, it will change, no, you don't have to worry - change will be negligible. You are replacing hydrogen peroxide with equimolar amount of water.

 

[sjb-2812]

http://en.wikipedia.org/w/index.php?title=Special:Cite&page=Hydrogen_peroxide&id=489569020 suggests that the density of hydrogen peroxide is significantly different to that of water; but as there is not much decomposition, this may be irrelevant.

 

[Borek]

http://en.wikipedia.org/w/index.php?title=Special:Cite&page=Hydrogen_peroxide&id=489569020 suggests that the density of hydrogen peroxide is significantly different to that of water; but as there is not much decomposition, this may be irrelevant.

Something else makes it completely irrelevant. He started with a 0.2M solution. Using density tables (or concentration calculator) you can easily check initial density to be around 1.0012 g/mL - so even after complete decomposition density change is just around 0.1%.