Question

. A student performed the experiment described in this module, using 5.00 ml of a 2.50% H2O2 solution with a density of 1.01 g ml-1. The water temperature was 24 degrees Celsius, and the barometric pressure in the laboratory was 30.50 in. Hg. After the student immersed the yeast in the peroxide solution, she collected 43.70 ml of O2. (A) convert the barometric pressure to torr. (B) Obtain the water vapor pressure at the water temperature. (C) Calculate the pressure, in torr, exerted by the collected O2 at the water temperature. (D) Convert the water temperature, in Celsius to kelvins. (E) Calculate the volume, in liters, that the collected O2 would occupy at STP. (F) Calculate the mass of the H2o2 solution. (G) Calculate the mass of H2O2 in 5.00 ml of the H2o2 solution. (H) Calculate the number of moles of H2O2 reacting. (I) Calculate the number of moles of collected O2. (J) Calculate the molar volume of O2 at STP. (K) calculate the percent error for the experiment.

Answer 1

(A) 1 in.Hg=25.40 torr (Conversion formula)

Therefore, 30.50 in.Hg=30.50*25.40 torr=774.7 torr (answer)

(B) The formula of calculating vapor pressure of water is

P=10^A-B/(C+T) where A,B,C are Antoine constants of water

When water temperature is in range 1-100 degree Celsius, A=8.07131, B=1730.63, C=233.426

When water temperature is in range 99-374 degree Celsius, A=8.14019, B=1810.94, C=244.485

Therefore, at water temperature 24 degree celsius,

P=10^{8.07131-1730.63/(233.426+24)}=10^{8.07131-6.7228}=10^1.3485=2975 Pa=2.975 KPa

(C) Using gas law, PV=nRT where R=universal constant=8.314 JK^-1mol^-1

Since 1 mol of O₂ is produced, n=1, T=24 C=297.15 K (conversion T(K)=T(C)+ 273.15), V=43.70mL=0.0437L

P=nRT/V= 1*8.314*297.15/0.0437

=56533.2975 Pa

=56533.2975*0.0075 torr

=423.9997 torr

=424 torr (answer)

(D) The conversion formula T(K)=T(^.C)+ 273.15

T(K)=24+273.15

=297.15(Answer)