Mass of Mg (g) Initial volume of Syringe (mL) Final volume of Syringe (mL) Volum
ID: 537864 • Letter: M
Question
Mass of Mg (g) Initial volume of Syringe (mL) Final volume of Syringe (mL) Volume of H_2 (mL) Barometric pressure (torr) Ambient temperature (degree C) Vapor pressure of H_2O (torr) Analysis Your goal is to determine the value of R from your data. Perform each of the following operate Attach a page showing your work. You do not have to show work for every trial, but show work for one sample trial for each step. 1. Determine the number of moles of magnesium consumed in the reaction, which will be equation the number of moles of H_2 produced (see Equation (2)). Moles of Mg (g) 2. Use Equation (2) to determine the pressure of H_2 in the sample. Express your answer atmospheres. Pressure of H_2 (atm) 3. Convert the ambient temperature to Kelvin. 4. Use Equation (1) to calculate R. Remember to convert mL to L to get the units below. 5. Calculate the average and standard deviation of your values of R. Average R (L atm/(mol K)) Standard deviation in R (L atm/(mol K))Explanation / Answer
trial-1 trial-2 trial-3
mass of Mg(g) 0.02 0.02 0.017
Atomic weight of Mg 24 24 24
moles= mass/atomic weight
moles of Mg 0.02/24= 0.000833 0.02/24=0.000833 0.017/24= 0.000708
moles of Hydrogen 0.000833 0.000833 0.000708 (same as mole of Mg) ml
volume of hydrogen (ml) =26-3.9= 22.1ml 24.6-3.9=20.7 22-3.9= 18.1
Volume of hydrogen (L)= 22.1/1000 =0.0221 20.7/1000=0.0207 18.1/1000=0.0181
Total pressure (mm Hg) 762.063 762.063 762.063
vapor pressure of water (mm Hg) 25.2 25.2 25.2
partial pressure of hydrogen =total pressure- saturation vapor pressure of hydrogen
Partial pressure of hydrogen (mm Hg) = 762.063-25.2= 736.863 ( same for all trials)
partial pressure of hydrogen in atm= partial pressure of hydrogen/760 = 736.863/760 atm =0.969 atm
from gas law PV=nRT
T= 26 deg.c= 26+273=299K
R= PV/nT, R will be in L.atm/mole.K
trial-1 trial-2 trial-3
R 0.969*0.0221/(299*0.000833) 0.969*0.0207/(299*0.000833) 0.969*0.0181/(299*0.000708)
0.08598 0.080534 0.070418
Average R=( 0.08598+0.080534+0.070418) /3 =0.0789 L/atm/mole.K