The following data were recorded in determining the molar mass of a volatile liq

Question

The following data were recorded in determining the molar mass of a volatile liquid following the Experimental Procedure of this experiment. Complete the table for analysis. (See Report Sheet.) Record calculated values with correct number of significant figures. 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, C and D. a. If the atmospheric pressure of the flask is assumed to be 760 torr in question 3, what is the reported molar mass of the vapor? b. What is the percent error caused by the error in the recording of the pressure of the vapor? %error = M_difference/M_actual times 100.

Explanation / Answer

Given

T = 98.7 C = 273 + 98.7 = 371.7 K

Volume V = 0.152 L

Pressure = 0.989 atm

R = 0.08206 L.atm/mol.K (data for R)

PV = nRT

n = PV / RT = 0.989 atm * 0.152 L / 0.08206 L..atm/mol.K * 371.7 K = 0.00493 moles Answer D1

mass of Vapour = 74.921 - 74.722 = 0.199 g Answer D2

Molar mass of vapour = Mass / No. of moles = 0.199 g / 0.00493 moles = 40.36 g/mol Answer D3

Given

46.5 g/mol and 43.1 g/mol

Average molar mass = 46.5 g/mol + 43.1 g/mol / 2 = 44.8 g/mol Answer 3 b a

Standard deviation = sqrt ( (1/2) * (( 46.5 - 44.8)2 + ( 43.1 - 44.8)2) ) = 1.7

relative standard deviation = (SD/ average) * 100 = (1.7/44.8) * 100 = 3.795 %

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