It was important that no water gets on the aluminum foil during the heating process. If water does get on the aluminum foil, it must be wiped off before reweighing your flask. A typical drop of water has an approximate volume of 0.05 mL and water has a density of 1.0 g/mL. Calculate the molecular weight of your unknown (using your data from the experiment) if 2 drops of water accidently remained on the aluminum foil when you reweighed your flask. First ExperimentSecond ExperimentWeight of dry apparatus, g67.520,91.297gWeight of apparatus pluscondensed vapor, g67.712991.620gWeight of condensed vapor, g0.29 2g0.3 239Total volume of flask, mL149.2mL152,5 mLTotal volume of flask, liters0.149 L0.1525LBarometric pressure, mm Hg752752Barometric pressure, atm0.9896.989Temperature of vapor, °C84.1c78.2°cTemperature of vapor, K357.25K351.35 KMolecular weight of liquid,g/mole58.05Gl,76Average molecular weightof unknown59.905g/molPV= nRT.989atm).494n(0.08206 aut in)(357.25k)n=0,00503 molPV=nRTCo.989atm)(0,15a5)=n(008460n) (351.351)Oi29agmw-igmol58.05n=0.00523mol1150.32390,00523mal0,00503 mol61.769/mol

Ideal gas equation gives the relation between 4 different parameters of air. The parameters pressure, temperature, volume and number of moles of the gas that are linked by the proportionality constant that is gas constant R. This equation can be used to find the fourth unknown property, if other three properties of the gas are known. The molecular weight of a substance is measure of mass of substance per mole quantity of the substance. In this experiment shows the connection between the Ideal gas equation and molecular weight of the substance. In the given experimental data, the volume of the flask, barometric pressure and the temperature of the vapor will remain the same. Therefore, the number of moles of the vapors will also remain same in both the experiments. The number of moles of the vapor in experiment 1 is 0.00503 mol and number of moles of vapor in experiment 2 is 0.00523 mol. Now consider that the flask contains 2 drops of water while reweighing. Each drop has volume of 0.05 mL and the density of water is 1 g/mL. Therefore, the mass of the water drops can be calculated by multiplying the number of water drops, volume of each drop and density of the water as follows. Mass of water drop =Number of drops×volume×density=2×0.05 mL×1.0 g/mL=0.1 g Since mass of water drop is 0.1 g, mass of condensed vapor in both the experiment will have additional 0.1 g. Now the weight of condensed vapor after water drop remained is calculated by adding the weight of the apparatus plus condensed vapor and weight of the water drops. Then, this weight is subtracted by the weight of dry apparatus. The weight of condensed vapor with water drop is calculated as follows. Wt of Condensed vapor+water drop=Wt of Apparatus +Condensed vapor+Wt ofwater drop-Wt of Dry Apparatus The weight for experiment 1 is calculated as follows. Wt of Condensed vapor+water drop=67.812 g+0.10 g-67.520 g=0.392 g Similarly, for the experiment 2, the weight is calculated as follows. Wt of Condensed vapor+water drop=91.620 g+0.10 g-91.297 g=0.423 g Now the molecular weight is calculated by dividing the new weight by the number of moles in each experiment. Molecular weight =New Weight of condensed vaporNumber of molesMolecular weight in experiment I =0.392 g0.00503 mol=77.93 g/mol Therefore, the molecular weight in experiment 1 is 77.93 g/mol Similarly, the molecular weight in experiment 2 is calculated as follows. Molecular weight in experiment I =0.423 g0.00523 mol=80.88 g/mol The molecular weight obtained in second experiment is 80.88 g/mol. The average molecular weight of unknown is calculated by adding the molecular weight of both the experiments and dividing the sum by 2. Avrage molecular weight =77.93 g/mol +80.88 g/mol2=79.41 g/mol Therefore, the average molecular weight of the unknown is 79.41 g/mol.