[integral0]

In an experiment which measures the volume of a generated gas by measuring the volume of displaced water, what volume of water (18.0152 g/mol) would be displaced by 250.0 ml of 3.00% by mass aqueous hydrogen peroxide (34.0146) at 19.5 deg C, on a day when the barometric pressure is 749.0 torr?  Assume the density of the hydrogen peroxide is 1.00 g/mol.  The vapor pressure of water at 19.5 deg C is 17.0 mm Hg.  The atomic weight of Oxygen is 15.9994 daltons.

2H202 -> 2H20 + 2O2

a. 2.47 ml

b. 2.76 x 10^3 ml

c. 2.74 x 10^3 ml

d. 2.64 x 10^3 ml

e. 5.52 ml

--------------------Anyone have any ideas?

I tried getting the grams of H2O2 and converting that to moles..and than to ml...and then I got stuck...b/c I don't know how to change it correctly into ml of water.

 

Thanks

[Demotivator]

ml water displaced refers to the volume of gas generated. The gas is comprised of O2  from the peroxide and water vapor from the water itself (note that the vapor pressure of water is given).  The pressure of the gases in the displacement have to equal the barometric pressure which is 749/760 = .9855 atm..

V = nRT/P
P = .9855 atm
R = .0821
T= 292.5 K

n is total moles of vapor and O2.
2H202 -> 2H20 + O2  (so O2 is 1/2 moles of H2O2)
n(O2) = (1/2)(.2205) = .1102 moles
need to find moles H2O vapor.

moles are proportional to partial pressure .
P(O2) = P(tot) - P(H2O) =  749 - 17 = 732 torr
hence,
n(H2O) = (17/732)* n(O2) = (17/732)* .1102 = .0026 moles
 total moles: n = .1102 + .0026 = .1128

now the volume can be calculated:
V = nRT/P
V = .1128(.082)(292.5)/.9855 = 2.74 liters = 2.74x10^3 ml