A student performed the experiment described in this module. The 5.00-mL mass of
ID: 895550 • Letter: A
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.
(a) Convert the barometric pressue to torr.
(b) Obtain the water vapor pressure at water temperature.
(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.
(l) Calculate the percent error for this determination.
Explanation / Answer
On Mixing Yeast the decomposition of H2O2 Takes place as follows :
H2O2(l) -----> H2O (l) + 1/2O2( g)
.108 gram
a) barometric Pressure = 31.2 in. Hg = 79.248 mm of Hg = 79.248 torr ( 1 torr = 1 mm of Hg)
b) The initial pressure is due to only water so vapour pressure of water = 79.248 torr at 296 k
c) PV=nRT
PO2 = nRT/V = (3.18x10-3x.0821x296/2)/(38.6x10-3) = 1 atomosphere = 760torr
assuming all the H2O2 decompsed
d) 1 atm = 760 torr
e) volume of O2 = 38.6 ml
f) Volume in litres = 38.6 x 10-3 litre
g) temperature in Kelvin = 23 + 273 =296 K
h ) Mass of H2O2 = 2.15% x 5.03 gram = .108 gram pure H2O2
i ) density of solution should be given to find mass of Solution
j) moles of H2O2 REACTED = .108/34 = 3.18x10-3
k ) R = .0821 lit/atm/molK
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.