6. In the molecular weight determination by the Dumas method, as ilustrated belo

Question

6. In the molecular weight determination by the Dumas method, as ilustrated below. several grams of an organic liquid were drawn into the bulb. The bulb was a water bath at 87°C and kept there until all the liquid was vaporized and the air in the bulb was replaced by the vapor. The tip was then sealed shut and the bulb removed from the bath, cooled, dried and weighed at room temperature. The weight of the bulb filled with the vapor was 36.127 g the weight of the empty (evacuated) bulb was 35.427 g:the bulb's volume was 200.0 mL; room temperature was 25 C; and the barometric pressure was 720 mm Hg. Calculate the molecular weight of the organic liquid then immersed in

Explanation / Answer

empty bulb = 35.427 grams

full bulb = 36.127 grams

mass of vapor = 36.127 - 35.427 = 0.7 grams of vapor

Volume = 200 ml or 0.2 L

pressure = 720 mmHg or 0.9473 atm

Temperature = 25C or 298.15 K

from the ideal gas equation

PV = n RT

p is pressure, v is volume, n is moles, R is gas constant, T is temperature in kelvin

moles = mass / molecular weight

PV = mass * R * T / molecular weight

molecular weight = mass R T / P V

molecular weight = 0.7 grams * 0.082 * 298.15 / (0.9473 atm * 0.2L)

molecular weight = 90.33 g/gmol

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