Interpretations of Racial Change

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EXPERIMENT 11 Molar Mass of a Volatile Liquid DATA Run | UNKNOW | Mass of flask and foil(g) | 90.085g | Mass of flask and foil and condensed vapor (g) | 90.640g | Temperature of boiling water (˚C ) | 98˚C | Barometric pressure (torr) | 761mm Hg | Volume of flask (mL) | 140 mL | RESULTS Mass of unknown ( condensed vapor) | 0.555g | Volume of flask (vapor)(L) | 0.14 L | Temperature of vapor(k) | 371K | Molar mass of unknown (g/mole) | 118.380 g/mole | Advance Study Assignment 1) How would each of the following errors affect the outcome of this experiment? Would it make the molar mass high or low? Give your reasoning in three sentences or less in each case. a) The hole in the aluminum foil was quite large. if you were to have a larger hole in your aluminum cap, you would be losing vapor to the room. If this is the case then you would have a falsely high weight of vapor. It seems that you take a known volume of vapor and condensed and weigh it b) Water vapors condensed under the aluminum foil before the final weighing. If Water vapors condensed under the aluminum foil before the final weighing, the final weight would be lower than the fact. There will have some error. 2) A volatile liquid was allowed to evaporate in a 43.298 g flask that has a total volume of 252 ml. the temperature of the water bath was 100˚C at the atmospheric pressure of 776 torr. The mass of the flask and condensed vapor was 44.173 g. calculate the molar mass of the liquid. T = 273 + 100 = 373 V = 252 mL = 1 L / 1000 mL = 0.252 L P = 776 Torr R= 0.0821 mass of 44.173 - 43.298 g = 0.875g moles of gas = PV / RT = 776 x .252 / 62.363 x (273+100) =0.00841 moles molar mass = 0.875g / 0.00841 moles = 104.1 g/

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