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 atmliter/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 atmliter/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
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '07
 Goo
 Amount of substance

Click to edit the document details