1. To find the mass of the vapor: Sample 1 Sample 2 25.25g -0.01 25.69g -0.01 25
ID: 498315 • Letter: 1
Question
1. To find the mass of the vapor: Sample 1 Sample 2 25.25g -0.01 25.69g -0.01 25.55g+-0.01 25.71gt-0.01 Mass of tube/cover Mass of tube/cover/vapor Mass of vapor 25.55-25.25 0.30 g +-0.01 25.71-25.69 0.02 g +-0.01 99 C +273.15 372.15 k 60 C+273.15 333.15 k Temperature of vapor Pressure of vapor 1.02 atm 1.02 atm 2. To find the volume of the tube 24 C+273.15 297.15 K Temperature of water in the tube 33.16g+-0.01 Mass of water 0.99732 g/ ml Density of water 33.16/0.99732 33.25 ml (0.03325 l) Volume of tube volume of the water 3. Results: Mole of vapor in the tube 1.11 10 -3 mole 1.23 10 -3 mole 270 g mole 16 g/mole Molar mass of vapor 9.0 0.39 g/l Density of vapor Mean value of molar mass of 270+16/2 143 g/mole vapor Mean value of density of vapor 9.0+0.59/2 5 g/lExplanation / Answer
1. There are many sources of error in this experiment. Few are listed below with answer d being the important reason:
a) The weighing balance used is not a precision balance. The readings should have been atleast upto 4 decimal place. Thus chances of error are high.
b) The cover might not be proper fitting thus resulting in the loss of gas
c) The gas might be lighter than air, and thus difficult to weigh.
d) The two samples are subjected to different temperatures and at constant pressure there will be a change in volume. The unaccounted change in volume leads to an error in measuring the molar mass.
2. % Error = | (Theoretical value - Experimental value) / Theoretical value | X 100
(NB: 58.08 g/mol is round off to 58 g/mol)
% Error = | 58 - 143/ 58| X 100 = 146 %
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