3. a. The data were recorded in determining the molar mass of a volatile liquid

Question

3. a. The data were recorded in determining the molar mass of a volatile liquid following the Experimental Procedure of this experiment. Complete the table for (See Report Sheet.) Record calculated the connect number of significant figures. Calculation Zone A. Preparing the sample I. Mass of dry nask, foil, and rubber band (R) 74.722 Part D. B. Vaporize the sample 2. Temperature of boiling water 987 3. Mass of dry flask, foi rubber band and vapor C. Determine the volume and Pressure Part D2 of the vapor 1. Volume of 125 mL flask (L) 2. Atmospheric pressure (orr atrn) 752, 0,989 D. Calculations 1. Moles of vapor, n v (mo) Equation 12.1 Part D.3 Show calculation. 2, Mass of vapor, m,mur (g) Show calculation. 3, Molar mass of vapor (g/mol Show calculation. 3. b. For Trials 2 and 3. the molar mass of the vapor was determined to be 46.5g/mol and 43.1 g/mol respectively. a. What is the average molar mass of the vapor? Data Analysis, B. b. What are the standard deviation and the relative standard deviation (%RSD) for the molar mass of the vapor? Data Analysis, Cand D.

Explanation / Answer

GIven

mass of flask = 74.722 g

Mass of flask, vapor = 74.921 g

Mass of vapor = 74.921 g - 74.722 g = 0.199 g Answer D2

Given

Volume = 0.152 L

Pressure = 0.989 atm

R = 0.08206 L.atm/mol.K Data

T = 98.7 C = 371.7 K

No. of moles n = PV / RT = 0.152 L * 0.989 atm / (0.08206 L.atm/mol.K * 371.7 K) = 0.00493 moles Answer D1

Molar mass of vapor = Mass / no. of moles = 0.199 g / 0.00493 moles = 40.38 g/mol Answer D3

3.b

Given

molar mass = 46.5 , 43.1

Average molar mass = (46.5 + 43.1) /2 = 44.8 Answer 3b.a

S.D = sqrt(1/2 ( (46.5-44.8)2 + (43.1-44.8)2 ) = 1.7 Answer 3b.b

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