Please complete what is missing from the First and Second determination. Third d

Question

Please complete what is missing from the First and Second determination. Third determination is not needed.

126 PRO P 481: Determining the Molar Mass of a Volatile Liquid by the Dumas Method name section date Data Sheet unknown identification code determination first second third mass of flask, boiling stone, foil cap, and unknown after cooling, g ,iS mass of flask, boiling stone, and foil cap, g 35.93 3322 13S.332 mass of unknown, g water bath temperature at complete vaporization, °C water bath temperature, K barometric pressure, in. Hg barometric pressure, atm volume of flask, mL volume of fiask, L density of vaporized unknown, gL- molar mass of unknown, g mol-1 average molar mass of unknown, g mol 1 accepted molar mass of unknown, g mo1 percent error, %

Explanation / Answer

For Determination First,

Mass of unknown = 136.562 - 135.332 = 1.23 g

Barometric Pressure in Hg = 760 mm

Barometric Pressure in atm = 1

Density of Unknown = 1.23 / 0.27 = 4.555 g/L

Moles of Unknown = PV / RT = (1x0.27) / (0.0821 x 370.15) = 0.008885

Molar Mass of Unknown = 1.23 / 0.008885 = 138.44 g/mol

For Determination Second,

Mass of unknown = 136.458 - 135.332 = 1.126 g

Barometric Pressure in Hg = 760 mm

Barometric Pressure in atm = 1

Density of Unknown = 1.126 / 0.27 = 4.1704 g/L

Moles of Unknown = PV / RT = (1x0.27) / (0.0821 x 370.15) = 0.008885

Molar Mass of Unknown = 1.126 / 0.008885 = 126.73 g/mol

Average Molar Mass of Unknown = (138.44+126.73) / 2 = 132.585 g/mol

Accepted Molar mass of Unknown = 84.16 g/mol

Percent Error = (132.585 - 84.16) / 84.16 x 100 = 57.54 %

Just try to put correct value of Pressure, you haven't given it so I have assumed it to be atmospheric.

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