Part 1: Information for Question PV=nRT In an experiment we reacted 0.0595g of m
ID: 1069269 • Letter: P
Question
Part 1: Information for Question
PV=nRT
In an experiment we reacted 0.0595g of magnesium in a eudiometer filled with water and HCL. The water temperature was 22.0 C. The barometer in the lab read 32.25in. Hg and 58.2 ml of gas was collected.
molar mass - 24.305 g
K = 273.15 + 22.0 = 295.15 K
58.2 ml/1000 = 0.0582 L
32.25 in times 25.4 mm Hg = 819.15
819.15 - 19.8 = 799.35 19.8 from the vapor pressure of water chart: 22.0 Celius = 19.8 pressure/mmHg
799.35 divided by 750 will get you atm: 1.052
PV=nRT. We finding R. P= 1.052 V= 0.0582 L n= 8.002448 T= 295.15
R = 0.0847
Question:
For each of the following experimental or calculation errors, tell if the calculated value of R [0.0847] would be too high, too low or unchanged explained answers.
A: some air bubbles were introduced when the gas buret was inverted.
B: the mass of the magnesium was recorded too low
C: some of the magnesium escaped from the copper wire basket and stuck to the inside of the gas buret remaining unreacted.
D: the temperature of the water was recorded too low E: the amount of HCl used was just right
Explanation / Answer
R = P V / n T
(A) Addition of air bubble scause increase in Pressure whihc inturn casuse increase in R value
(B) Measuring of mass of Mg too low would cause calculation of moles of H2 also low. Hence R value would be higher.
(C) It cause the release of less amount of H2 which cause low pressure and volume of H2 which inturn cause low R value
(D) Low T value gives low R value.