Question

i need part 2 and part 3 to be solved thanks please please urgent

Instructor Name: Student Name: DATA (EXP #13): Part 1: The Action of Atmospheric Pressure Observations upon immersing the can in ice-water bath: Part 2: Determination of the Molar Mass of Gases Barometric pressure (mmHg): 734.2 mm Room temperature (0 23, 6"c CO2 Natural Gas Mass of flask/stopper filled with air (g) 126 069 125.959 la 0146-16 146.9 Volume of water used to fill flask (mL) 267-nL 56 mL Part 3: Determination of the Molar Mass of an Unknown Liquid Heat source temperature (°C) Trial 1 Trial 2 d1:62 1:670 18.1C 98.9 795 Mass of test tube and foll cover (g) 7.9626 Temperature of heat source (°C) Mass of test tube and foil and unknown sample Mass or volume of water used to fill test tube mL) CHE-1251 Experiment 13 Student Name: Instructor Name: CALCULATIONS: Part 2: Determination of the Molar Mass of a Gas Barometric pressure (mmHg) Room Temperature (°C): Using the lab values for Pressure and temperature, calculate the density of dry air (similar to your pre-lab assignment), show work for your calculation below CO2 Natural Gas Mass of air in flask (9) (show work below) Mass of empty flask (g) (show work below) Mass gas in flask (g) show work below) Molar Mass of Gas (g/mol) (show work below) % Error (show work below) CHE-1251 Experiment 13 14 Instructor Name Student Name: Part 3: Determination of the Molar Mass of an Unknown Liquid Heat source temperature (C) Trial 2 Trial 1 Mass of vapor sample (g) (show work below) Number moles vapor (mol) (show work below) Molar mass (g/mol) (show work below) Average molar mass (g/mol) 1. Based on the molar mass, its boiling point and the values on the NFPA label, propose an identity for your unknown liquid. Describe the tasks you performed as part of your service learning project (part 4). 2. CHE-1251 Experiment 13 15

Answer 1

1) Carbon dioxide :

Mass of CO₂, m = (Mass of flask filled with gas - Mass of flask filled with air) = 146.96 g - 146.77 g = 0.19 g

Volume of water , V = 267 ml = 0.267 lts

Barometric Pressure , Patm = 734.2 mmHg

Vapor pressure of water at 23.6 C, Pwater = 21.848 mmHg

Pressure of dry carbon dioxide = Patm -Pwater = 734.2 mmHg - 21.848 mmHg = 712.352 mmHg = 712.352 /760 = 0.937 atm

Ideal gas constant, R = 0.0821 lt atm/mol K

Temperature , T = 23.6 C +273 = 296.6 K

PV = nRT

0.937 atm x 0.267 lts = m/MW) x 0.0821 lt atm/mol K x 296.6 K

0.250 = (0.19 g/MW) x 24.351/mol

0.250 mol/24.351 = 0.19 g/MW

0.01027 mol = 0.19 g/MW

MW = 0.19 g/0.01027 = 18.5 g/mol

Theoretical molar mass of CO₂ = 44 g/mol

Percent error = [exp - theoretical]/theoretical) x 100= [18.5 -44]/44) x 100 = 57.95 %

2) Natural gas

Mass of natural gas , m = (Mass of flask filled with gas - Mass of flask filled with air) = 126.06 g - 125.95 g = 0.11 g

Volume of water , V = 256 ml = 0.256 lts

Barometric Pressure , Patm = 734.2 mmHg

Vapor pressure of water at 23.6 C, Pwater = 21.848 mmHg

Pressure of dry carbon dioxide = Patm -Pwater = 734.2 mmHg - 21.848 mmHg = 712.352 mmHg = 712.352 /760 = 0.937 atm

Ideal gas constant, R = 0.0821 lt atm/mol K

Temperature , T = 23.6 C +273 = 296.6 K

PV = nRT

0.937 atm x 0.256 lts = m/MW) x 0.0821 lt atm/mol K x 296.6 K

0.2399 = (0.11 g/MW) x 24.351/mol

0.2399 mol/24.351 = 0.11 g/MW

0.00985 mol = 0.11 g/MW

MW = 0.11 g/0.00985 = 11.17 g/mol

Theoretical molar mass of CO₂ = 19 g/mol

Percent error = [exp - theoretical]/theoretical) x 100= [11.17 -19]/19) x 100 = 41.21 %

c) Unknown liquid :

Trial 1 :

Mass of unknown gas , m = (Mass of flask filled with gas - Mass of flask filled with air) = 8.378 g - 7.962 g = 0.416 g

Volume of water , V = 8.9 ml = 0.0089 lts

Vapor pressure of water at 98.7 C, Pwater = 727.02 mmHg = 0.957 atm

Ideal gas constant, R = 0.0821 lt atm/mol K

Temperature , T = 98.7 +273 = 371.7 K

PV = nRT

0.957 atm x 0.0089 lts = m/MW) x 0.0821 lt atm/mol K x 371.7 K

0.00852 = (0.416 g/MW) x 30.52/mol

0.00852 mol/30.52 = 0.416 g/MW

0.00028 mol = 0.416 g/MW

MW = 0.416 g/0.00028 = 1485.7 g/mol

Trial 2 :

Mass of unknown gas , m = (Mass of flask filled with gas - Mass of flask filled with air) = 7.752 g - 7.67 g = 0.082 g

Volume of water , V = 8.9 ml = 0.0089 lts

Vapor pressure of water at 98.9 C, Pwater = 730 mmHg = 0.961 atm

Ideal gas constant, R = 0.0821 lt atm/mol K

Temperature , T = 98.9 +273 = 371.9 K

PV = nRT

0.961 atm x 0.0089 lts = m/MW) x 0.0821 lt atm/mol K x 371.9 K

0.00852 = (0.082 g/MW) x 30.53/mol

0.008552 mol/30.53 = 0.082 g/MW

0.00028 mol = 0.082 g/MW

MW = 0.082 g/0.00028 = 29.286 g/mol

Average molar mass = 29.26 g/mol + 1485.7 g/mol)/2 = 757.5 g/mol