Gas Law Chem Lab (Two Questions) 1.) Through the application of Dalton’s Law and

Question

Gas Law Chem Lab (Two Questions)

1.) Through the application of Dalton’s Law and the Combined Gas Law, calculate the molar volume of hydrogen for each of your trials in the previous experiment. Calculate the average, standard deviation, and relative standard deviation for your experiment. If the actual value for molar volume should be 22.414 L/mol, calculate the percent error using your average molar volume as the experimental value.

2.) Using the information found in Question 1 as well as the ideal gas law, calculate your experimental value for the ideal gas constant, R, for each of your trials. Calculate the average, standard deviation, and relative standard deviation for your experiment. If the actual value for R should be 0.08206 (L atm)/(mol K), calculate the percent error using your average ideal gas constant as the experimental value.

This Postiab is combined with Redox... you may want to look ahead to see what is required for that Postlab and start planning for that deliverable. This Data Table may help you as you start to compile that information. PostLab: Data Table Trial 3 Trial 1 Trial 2 Atmospheric pressure (mm Hg) O. 2 Mass of Mg metal (g) Volume of 3 M HCI (mL) n Volume of H2 gas (mL.) Water temperature (C) T m PC Partial pressure of water vapor (mm Hg; from Table 1) Difference in water levels (mm) 2-1

Explanation / Answer

Moles of Magnesium used = 0.029/24.305 = 0.0012 moles (average of the H2 produced in three trials is 0.029gm)

Barometric = total pressure (760 mmHg)           760 mm Hgtotal – 17.54 mmHg H2O = 742.46mmHg H2

The average of the three cases that for the amount of H2 gas produced is 24.8 mL.(for the three trials) = 0.0248 L

Use combined gas law to convert gas volumes to those at STP.

P1V1/T1 = P2V2/T2 let STP be 2nd conditions      V2 = P1V1T2/T1P2

                                                      Vstp = (742 46mmHg) (0.0248L) (273 K)

                                                                           (293 K) ( 760 mmHg)                 =   0.02257 L

Divide volume at STP by the number of moles to get an experimentally determined molar volume at STP.      0.02257 L/0.0012moles     = 18.80 L/mole.

Percentage error = (22.4-18.8)/22.4 *100 = 16.07%.

2). PV= nRT P=pressure, V=Volume ,n= no.of moles, R = Universal Gas constant)

R = PV/nT = 742.46 * 0.2257 L/ 0.0012 mole* 293K*760 = 0.627 atm lit deg-1 mol-1

The percentage error = 0.821-0.627 / 0.821 * 100 = 23.6%.

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