2. A student performed the experiment described in this module, using 5.00 ml of

Question

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.

Explanation / Answer

(a) Given the pressure, P = 30.50 inch Hg = 30.50 inch x ( 2.54 cm / 1 inch) = 77.47 cm Hg

= 77.47 cm Hg x(10 torr Hg/ 1 cmHg) = 774.7 torr Hg (answer)

(b) Water vapor pressure = Total pressure - Atmospheric pressure = 774.7 torr Hg - 760 torr Hg = 14.7 torr Hg

(c) Mass of H2O2 solution = Volume of H2O2 solution x density = 5.00 mL x (1.01 g/mL) = 5.05 g

Also given the solution is 2.5% by weight

Hence mass of H2O2 in the solution = 5.05 g x(2.5 / 100) = 0.12625 g

Moles of H2O2 = mass / molecular mass = 0.12625g / 34 g/mol = 0.00371 mol

The balanced chemical reaction for the decomposition of H2O2 is

2H2O2 ---- > 2H2O + O2

2 mol 2 mol 1 mol

2 mol of H2O2 produces 1 mol O2

Hence 0.00371 mol of H2O2 that will prduce the amount of O2 = 0.00371 mol H2O2 x ( 1 mol O2 / 2 mol H2O2)

= 0.00186 mol O2

Given volume of O2 collected, V = 43.70 mL = 0.04370 L

Temperature, T = 24 DegC = 24 + 273 = 297 K

Applying ideal gas equation

PV = nRT

=>P = nRT / V = (0.00186 mol x0.0821L.atm.mol-1K-1 x 297K) /  0.04370 L = 1.038 atm

= 1.038 atm x (760 torr / 1 atm) = 789 torr (answer)

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