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MASC 110 Solution HW3

# MASC 110 Solution HW3 - University of Southern California...

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University of Southern California MASC 110L Solution Set #3 E. Goo 1. Calculate the volume occupied at 25 o C and 1 atm pressure by the gas evolved from 1 cm 3 of solid carbon dioxide(density 1.56 g cm -3 ). Pressure is one atm and temperature is 25 + 273.15 = 298.15 K One cm 3 of solid carbon dioxide has a mass of 1.56 g. The molar mass of carbon dioxide is 44 g/mole. Therefore one cm 3 of carbon dioxide has (1.56 g)/(44 g/mole) = 3.545 x 10 -2 moles. From the ideal gas law V = nRT/P = (3.545 x 10 -2 moles)(0.082 atm-liter/mole- K)(298.15 K)/(1 atm) = 0.8667 liters 2. The density of helium at 0 o C and 1 atm is 0.1786 g l -1 . Calculate its density at 100 o C and 200 atm. Pressure is initially one atm and temperature is 273.15 K Pressure is raised to 200 atm and temperature to 100 + 273.15 = 373.15 K Density is proportional to moles/volume=n/V From the ideal gas law n/V=P/RT therefore the density is proportional to P/T Initially P/T = 1/273.15= 3.66x10 -3 Finally P/T = 200/373.15=0.536. Therefore the density will increase to 0.1786x(0.536/3.66x10 -3 )=26.2 g l -1 . 3. In 1811 Avogadro assigned the formula C 10 H 16 O to camphor, on the basis of its vapor density and chemical analysis. The observed density of camphor vapor at 210 o C and 1 atm is 3.84 g l -1 . To what value of the molecular weight does this density correspond? Pressure is one atm temperature is 210 + 273.15 = 483.15 K. Consider a volume of one liter. From the ideal gas law n = PV/RT = (1 atm)(1 liter)/(0.082 atm-liter/mole- K)(483.15 K) = 2.52 x 10 -2 moles. From the density this many moles has a mass of 3.84 g. Therefore the molar mass is 3.84 g/2.52 x 10 -2 moles = 152.38 g/mole. This is also the molecular weight of 152.38 amu. Where one amu(atomic mass

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MASC 110 Solution HW3 - University of Southern California...

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