Experiment 18 Report Sheet Molar Mass of a Volatile Liquid Date lab Sec. Name- D

Question

Experiment 18 Report Sheet Molar Mass of a Volatile Liquid Date lab Sec. Name- Desk No. Unknown Number Trial I* Trial 2 Trial 3 an203 1203 2128 1. Mass of dry flask, foil, and rubber band () 2. Temperature of boiling water (°C, K) 3. Mass of dry flask, foil, rubber band, and vapor (g) 4. Volume of 125-ml flask (L) 21.41681529 87.468 935 atm 5. Atmospheric pressure (torr, atm) Calculations 1. Amount of vapor, np (mol) 2. Mass of vapor, m,or (g) 3. Molar mass of compound (g/mol) 4. Average molar mass (g/mol) Calculation of Trial 1. Show work here. Class Data/Group Molar mass Sample Unknown No. Experiment 18 225

Explanation / Answer

Trial 1 :
Mass of vapor, mvapor = 87.476-87.203 = 0.273 gm
P = 0.985 atm ; T = 94 + 273.15 = 367.15 K ; V = 125 mL = 0.125 L
PV = nRT ; R = 0.0821 atm L mol-1K-1
0.985 * 0.125 = n * 0.0821 * 367.15
nvapor = 4.08469*10-3 mol
nvapor = mvapor / Molar mass
4.08469*10-3 = 0.273 gm /Molar mass
Molar mass of the compound =0.273 / 4.08469*10-3 = 66.8348 gm/mol
Trial 2 :
mvapor = 87.529 - 87.203 = 0.326 gm
T = 96+273.15 = 369.15 K
0.985 * 0.125 = n * 0.0821*369.15
nvapor = 4.0625*10-3 mol
0.326 gm / 4.0625*10-3 mol = 80.2448 gm/mol
Trial 3 :
mvapor = 87.468 - 87.203 = 0.265 gm
T = 369.15 K
0.985 * 0.125 = n * 0.0821 * 369.15
nvapor = 4.0625*10-3 mol
0.265 gm / 4.0625*10-3 mol = 65.2297 gm/mol
Average Molar mass = (trial1 + trial2 + trial3) / 3
Avg Molar mass = (66.8348 + 80.2448 + 65.2297) / 3 = 70.7697 gm/mol

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