Question

A student performed the experiment described in this module. The 5.00-mL mass of the 2.15% percent by mass H2O2 solution used was 5.03 g. The water temperature was 23 C, and the barometric pressure was 31.2 in. Hg. After the student immersed the yeast in the peroxide solution, she observed a 38.60-mL volume change in system volume.

(c) Calculate the pressure, in torr, exerted by the collected O2 at the water temperature.

(d) Covert this pressue, in torr, to atmospheres.

(e) Calculate the volume, in milliliters, of collected O2.

(f) Convert this volume, in milliliters, to liters.

(g) Convert the water temperature, in Celsius, to kelvins.

(h) Calculate the mass of 5.00 mL of the H2O2 solution.

(i) Calculate the mass of H2O2 in 5.00 mL of the solution.

(j) Calculate the number of moles of H202 reacting.

(k) Calculate the number of moles of collected O2.

(l) Determine the proportionality constant, R, in L atm mol-1 K-1.

(m) Calculate the percent error for this determination.

please, please please show all work

Answer 1

(c) 1 in of Hg = 25.4 torr so 32.1 in Hg = 32.1*25.4 = 815.3401 torr = 815.3401 mmHg
P_total = P_O2 + P_H2o
where P_H2o is the vapor pressure of water at 23^oC =21.068 mmHg
815.3401 mmHg = P_O2 + 21.068 mm Hg
P_O2 = 815.3401 - 21.068 = 794.2721 mmHg

(d) 1 mm Hg = 1 torr so, 794.2721 mmHg = 794.2721 torr
1 atm = 760 torr so, 794.2721 / 760 = 1.04509 atm

(e) Volume change observed will be the answer i.e) 38.6 mL of O₂
(f) 0.0386 L of O₂ = 38.6/1000

(g) K = ^oC + 273.15 K So, 273.15+23 = 296.15 K