2. 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.
mass of solution of 5 ml = 5*1.01= 5.05 gms
H2O2 in the solution = 5.05*2.5/100=0.12625 gms
Moles of H2O2= 0.12625/34 ( Molecular weight of H2O2)= 0.003713 moles
Pressure =30.5 in Hg =30.5*25.4 mm Hg= 774.7 mm Hg= 774.7 torr
Vapor pressure at 24 deg.c = 22.4 mm Hg
Temperature =24 deg.c =24+273.15K= 297.15K
Volume of oxygen collected (V1)= 43.7ml at T1 =297.15 K and P1 =774.7 torr
P2= 760 Torr T2= 273.15 K (STP conditions)
from P1V1/T1= P2V2/T2
V2= P1V1*T2/(P2*T1)= 774.7*43.7*273.15/(760*297.15)=40.94 ml =0.04094L
1 mol of any gas at STP occupes 22.4 liter
0.04094 L occupies 0.04094/22.4 moles=0.001828 moles
Molar volume =moles/ Volume= 0.001828 Mol/ 0.04094 L=0.044643 Mol/L
Oxygen collected ( from PV= nRT
n= number of moles of oxygen =(774.7/760)*(43.7/1000)/ (0.08206*297.15)=0.001827 moles
error =(0.01828 ( from STP ) -0.001827)/0.01828 =0.0649%
Get Answers For Free
Most questions answered within 1 hours.