Name: _________________________________________
Homework 10 – FOR PRACTICE ONLY!! DO NOT TURN IN!!!
INSERT SPACES IF YOU WANT TO SOLVE THE PROBLEMS ON THIS DOCUMENT
1.) Explain why density would be proportional to molar mass.
Density is grams/mL while molar mass is grams/mole. Both of them have grams in the
numerator, so if the grams increase, the density goes up, if the grams increase, the molar mass
increases. Also, the more grams you have, the more mass you have and the more mass you
have, the more stuff present, so the more dense the material becomes (think of rocks compared
to foam!)
2.) How would the effusion rate for He compare to the effusion rate of H2? Which would effuse faster?
Why?
Hydrogen (H2) has a molar mass of 2.02 grams/mol while He has a molar mass of 4.00
grams/mole. The effusion rate law states that the rate is INVERSELY proportional to the molar
mass. Thus, small molar masses have larger/fast effusion rates while larger molar mass have
lower/slower effusion rates. Since the molar mass of H2 is ½ the molar mass of He, it should be
expected that H2 should effuse TWICE as fast as He
3.) At a set pressure and temperature it takes 5 minutes for a 1.5 L sample of He to effuse through a
porous membrane. How long does it take a sample for 1.5 L of F2 to effuse under the same
conditions. (HINT: think about which species should effuse faster before you calculate anything!!)
mm He = 4.00 grams/mol
mm F2 = 38.00 grams/mol
This means that he should effuse MUCH faster than F2, or F2 should effuse slower than He (doesn’t
matter how you look at it as long as you know which one is faster (He) and which one moves slower (F2)
rateF 2
rateHe

mm He
mm F2
The 5 minutes given in the problem is not a rate, so we can’t plug that in for the rate of He!
rateF 2
4.00

rateHe
38.00
rateF 2
rateHe
 0.333
So let’s think about this, the ratio of rates of F2 to He is 1/3, that means that He effuses 3 times faster than
F2. That means that F2 effuses 3 times slower than He. It will take 33% MORE time for F2 to effuse than
for He.
If it takes 5 minutes for He to effuse, and it takes 3 times longer for F2 to effuse, then it must take 15
minutes for F2 to effuse. (3 x 5 = 15!) (or (F2 time)(0.333) = 5 therefore x = 15)
4.) The volume of 1.00 mole of ammonia gas at 1.00 atm of pressure was gradually decreased from 78
mL to 39 mL. What was the final pressure of ammonia if there was no change in temperature?
Make a list of the variables moles and temperature are held constant, volume and pressure are
studied.
P1 V1 P2 V2

n 1 T1 n 2 T2
Since n1 = n2 and T1 = T2 this expression reduces to
P1V1 = P2V2
P1 = 1.00 atm
P2 = ???
V1 = 78 mL
V2 = 39 mL
P2 =
P1 V1
(1.00 atm)(78 mL)
=
=
V2
39 mL
2.00 atm
5.) The pressure on a sample of ideal gas was increased from 715 mm Hg to 3.55 atm at a constant
temperature. If the initial volume of the gas was 48.5 mL, what was the final volume of the gas?
Make a list of the variables moles and temperature are held constant, volume and pressure are
studied.
P1 V1 P2 V2

n 1 T1 n 2 T2
Since n1 = n2 and T1 = T2 this expression reduces to
Since n1 = n2 and T1 = T2 this expression reduces to
P1V1 = P2V2
P1 = 715 mm Hg x
1 atm
= 0.941 atm
760 mm Hg
P2 = 3.55 atm
V1 = 48.5 mL
V2 = ????
V2 =
P1 V1
(0.941atm)(48.5mL)
=
=
P2
3.55atm
12.9 mL
6.) How many moles of air must be in a bicycle tire with a volume of 2.36 L if it has an internal pressure
of 6.8 atm at 17.0oC?
Only single values are given for P, V, T, and/or n so we will not be using the combined gas law.
We will be using the ideal gas law PV = nRT
P = 6.8 atm
V = 2.36 L
T = 17.0oC + 273.15= 290.2 K
n = ????
PV = nRT
n=
PV
=
RT
(6.8 atm) (2.36 L)
=
 2 Latm
(8.206 x 10
)(290.2 K)
molK
0.67 moles
7.) Hydrogen holds promise as an environmentally friendly fuel. How many grams of H2 gas are
present in a 50.0 L fuel tank at a pressure of 2850 psi at 20oC? (14.7 psi = 1 atm)
This is a practice problem where you are supposed to remember that certain variables are
“hidden” or can be determined using the PV = nRT expression. Let’s also examine what we
have in terms of variables given in the problem
P = 2850 psi x
1 atm
=194 atm
14.7 psi
V = 50.0 L
T = 20oC + 273.15 = 293 K
mm =
grams
grams
rearrange that and you have moles =
we know/can determine the mm of H2
moles
mm
H2: 2 x 1.01 = 2.02 grams/mol
PV = nRT
PV =
grams =
PVmm
=
RT
grams
RT
mm
(194 atm)(50.0L)(2.02
(8.206x10 2
grams
)
mole =
Latm
)(293 K)
molK
815 grams
8.) How does the density of a gas sample change when
a. Its pressure is increased
b. Its temperature is decreased
If the pressure is increased, then the molecules are forced closer together. There would be more
mass in the same space because the molecules are closer. If there is more mass in the same
space, then the density increased. Think about this. In a 1.00 L container, let’s just say there are
10 molecules per liter. Those 10 molecules have mass. If we increase the pressure on the gas,
then we will have more molecules present per liter, so more mass so the density increases!
Also, if you increase the pressure, chances are you are doing so by decreasing the volume –
since d = m/v, if the volume decreases, the density increases (they are inversely related to one
another!)
If the temperature is decreased, the molecules move slower and with less “gusto”. They are not
moving around as much, which means that more molecules will be present per some unit
volume. If more molecules are present (they aren’t moving a lot/away from one another), then
the measured density will be greater as well.
9.) Calculate the density, in grams per liter of O2 gas at 0oC and 1.00 atm
This is a practice problem where you are supposed to remember that certain variables are
“hidden” or can be determined using the PV = nRT expression. Let’s also examine what we
have in terms of variables given in the problem
P = 1.00 atm
V = ???
T = 0oC + 273.15 = 273 K
n = ???
It seems that we are missing too many things!! But remember, those “hidden” variables are in there.
And density is grams/Volume - so we can be missing two variables and we will call that density!
mm =
grams
grams
rearrange that and you have moles =
we know/can determine the mm of H2
moles
mm
H2: 2 x 1.01 = 2.02 grams/mol
PV = nRT
PV =
grams
RT
mm
grams
Pmm grams
and
= density (d)

V
RT
V
Pmm
d
RT
grams O 2
)
1 mole
=
Latm
)(273 K)
molK
1.00 atm (32.00
d=
(8.206 x 10 - 2
1.43
grams O 2
L
10.) Calculate the molecular weight of butane if 0.5813 grams of the gas fills a 250.0 mL flask at a
temperature of 24.4oC and a pressure of 742.6 mmHg.
This is a practice problem where you are supposed to remember that certain variables are
“hidden” or can be determined using the PV = nRT expression. Let’s also examine what we
have in terms of variables given in the problem
P = 742.6 mm Hg x
1 atm
=0.9771 atm
760 mm Hg
V = 250.0 mL = 0.2500 L
T = 24.4oC + 273.15 = 297.6 K
mm =
grams
grams
rearrange that and you have moles =
we know/can determine the mm of H2
moles
mm
H2: 2 x 1.01 = 2.02 grams/mol
PV = nRT
PV =
(grams)RT
mm =
=
PV
grams
RT
mm
Latm
)(297.6K )
molK
=
(0.9771atm)(0.2500 L)
(0.5813 grams)(8.2 06x10 2
58.11
grams
mol
11.) A sample of NH3 gas fills a 27.0 L container at -15oC and 2.58 atm. Calculate the volume of the gas at
21oC and 751 mmHg.
V1 = 27.0 L
T1 = -15oC + 273.15 = 258 K
P1 = 2.58 atm
V2 = ???
T2 = 21oC + 273.15 = 294 K
P2 = 751 mm Hg x
1 atm
= 0.988 atm
760 mm Hg
The number of moles stays constant! n1 = n2
P1 V1 P2 V2

n 1 T1 n 2 T2
Therefore, since n stays constant, the equation becomes
P1 V1 P2 V2

T1
T2
Solving for V2 gives us
V2 =
P1 V1 T2
(2.58atm)(27.0L)(294 K)
=
=
T1 P2
(258 K)(0.988atm)
80.3 L
12.) A tank contains 480.0 grams of oxygen and 80.00 grams of helium at a total pressure of 7.00
atmospheres. Calculate the following.
a) How many moles of O2 are in the tank?
b) How many moles of He are in the tank?
c) Total moles of gas in tank
d) Mole fraction of O2
e) Mole fraction of He
f) Partial pressure of O2
g) Partial pressure of He
480.0 grams O2 x
1mole O 2
= 15.00 moles O2
32.00gramsO 2
80.00 grams He x
1mole He
= 20.0 moles He
4.00gramsHe
Total moles = 15.00 + 20.0 = 35.0 moles total
XO2 =
moles O 2
15.00moles O 2
= 0.4285

total moles 35.0 total moles
XHe =
moles He
20.0 moles He
= 0.571

total moles 35.0 total moles
PO2 = XO2 x PT
PO2 = 0.4285(7.00 atm)
PO2 = 3.00 atm
PHe = XHe x PT
PHe = 0.571(7.00 atm)
PHe = 4.00 atm
And look, PO2 + PHe = 7.00 atm!!
13.) 96.0 grams of a gas occupies 48.0 L at 700.0 mmHg and 20.0 °C. What is its molecular weight?
This is a practice problem where you are supposed to remember that certain variables are
“hidden” or can be determined using the PV = nRT expression. Let’s also examine what we
have in terms of variables given in the problem
P = 700.0 mm Hg x
1 atm
=0.9211 atm
760 mm Hg
V = 48.0 L
T = 20.0oC + 273.15 = 293.2 K
mm =
grams
grams
rearrange that and you have moles =
we know/can determine the mm of H2
moles
mm
H2: 2 x 1.01 = 2.02 grams/mol
PV = nRT
PV =
(grams)RT
mm =
=
PV
grams
RT
mm
Latm
)(293.2K)
molK
=
(0.9211atm)(48.0L)
(96.0 grams)(8.2 06x10 2
52.2
grams
mol
14.) Calculate the total pressure of a mixture that contains 1.00 g H2 and 1.00 g of He gases in a 5.00 L
container at 21oC.
PT = PH2 + PHe
1.00 g H2 x
1.00 g He x
1mole H 2
= 0.495 moles H2
2.02 grams H 2
1mole He
4.00gramsHe
= 0.250 moles He
T = 21oC + 273.15 = 294 K
PV = nRT
n RT
PH2 = H2
=
V
n RT
PHe = He
=
V
(0.495moles H 2 )(8.206x 10 - 2
5.00 L
(0.250 moles He)(8.206 x 10 - 2
5.00 L
Latm
)(294 K)
molK
= 2.39 atm
Latm
)(294 K)
molK
= 1.21 atm
PT = PH2 + PHe
PT = 2.39 atm + 1.21 atm =
3.60 atm
15.) A mixture of 18.0 grams of hydrogen, 184.0 grams of helium, and 4.0 moles of hydrogen chloride are
placed in a flask. When the partial pressure of the helium is 178.00 mm Hg, what is the total
pressure in the flask? (10 pts)
This problem seems like there just isn’t enough info to solve it. The temperature isn’t given,
the volume isn’t given. We have some moles listed, and some grams, but then we are asked
about pressure. How in the world can we find total pressures if we don’t have V and T to get
the partial pressures of all the species so we can add the partial pressures together??? Well, we
don’t need that!
XA =
PA partial pressure of gas A

PT
total pressure
We are given, in the problem, the partial pressure of helium. If we can find the mole fraction of
helium, we will have 2 of the 3 variables in the equation above so we can solve for PT.
XHe =
moles He
total moles
So we need total moles – REMEMBER YOUR DIATOMICS!!!
18.0 g H2 x
1mole H 2
= 8.91 moles H2
2.02 grams H 2
184.0 g He x
1mole He
= 46.0 moles He
4.00 grams He
And there are 4.0 moles of HCl
Therefore, the total number of moles = 8.91 + 46.0 + 4.0 = 58.9 moles
XHe =
46.0 moles He
= 0.781
58.9 total moles
XA =
PT =
PA
PT
PA 178.00mm Hg
=

XA
0.781
227.9 mm Hg
Notice that the total pressure is > than the partial pressure – it should be since we have 2 other
gases in there!