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.

(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

Ans. #h. Mass of 5.00 mL of the H₂O₂ solution = 5.03 g [Given]

#i. Mass of H₂O₂ in 5.0 mL = 2.15 % (by mass) of solution

                                                = 2.15 % of 5.03 g

                                                = 0.108145 g

#j. Moles of H₂O₂ reacted (assuming all H₂O₂ reacts) = Mass of H₂O₂ / Molar mass

                                                = 0.108145 g / (34.01468 g/ mol)

                                                = 0.0031794 mol

#k. Balanced reaction of O₂ formation:                2 H₂O₂ ----> H₂O + O₂

Stoichiometry: 2 moles of H₂O₂ produces 1 mol O₂.

So,

            Moles of O₂ formed = (1/2) x Moles of H₂O₂

                                                = (1/2) x 0.0031794 mol

                                                = 0.0015897 mol

#l. Ideal gas equation:         PV = nRT      - equation 1

            Where, P = pressure in atm

            V = volume in L                   

            n = number of moles

            R = universal gas constant= 0.0821 atm L mol^-1K^-1

            T = absolute temperature (in K) = (⁰C + 273.15) K

Given,

            Moles of O₂ formed = 0.0015897 mol

Temperature, T = 296.15 K

            Pressure = 31.2 in Hg = 792.48 mm Hg = 1.0427 atm               ; [1 inch = 25.4 mm]

            Volume of O₂ formed, V = 38.60 mL = 0.03860 L                       ; [1 L = 10³ mL]

Putting the values, except that of R (to be calculated) in above equation-

            1.0427 atm x 0.03860 L = 0.0015897 mol x R x 296.15 K

            Or, 0.04205 atm L = (0.47079 mol K) R

            Or, R = 0.04205 atm L / (0.47079 mol K) = 0.0855 atm L mol^-1 K^-1

            Hence, R = 0.0855 atm L mol^-1 K^-1

#m. Error in value = Actual value – Calculated value

                                    = 0.0855 atm L mol^-1 K^-1 - 0.0821 atm L mol^-1 K^-1

                                    = 0.0034 atm L mol^-1 K^-1

% error = (Error in value / Actual value) x 100

                        = (0.0034 atm L mol^-1 K^-1 / 0.0821 atm L mol^-1 K^-1) x 100

                        = 4.14 %