Example 7.3

What is the molar concentration of fluorine in a 4.0-L flask that contains 4.2 g of fluorine? Use Equation 7.8, which shows that the molar

concentration of fluorine is equal to the

number of moles of fluorine divided by the vo

lume of the container. Thus, we start by

determining the number of moles of fluor

ine. Fluorine is a diatomic gas (F

), so M 2

= m

2(19.0) = 38.0 g

.mol

-1 and the number of moles is

n = 4.2 g F

× 2

1 mol F

2

38.0 g F

= 0.11 mol F 2

(^2)
Now, we can determine the concentration of F
with Equation 7.8. 2
[F
] = 2
0.11 mol F
2
4.0 L
= 0.028 mol
-1⋅L
or 0.028 M
The concentration of fluorine in the c
ontainer is 0.028 M (read 0.028 molar).
Example 7.4
0.12 mol Ne and 0.19 mol He are placed in a 6.0-L flask at 55
oC. What is the total
pressure in the flask, and what are the partial pressures of the two gases? We first determine the two partial pressures wi
th the ideal gas law. The temperature of
the mixture is T = 55 + 273 = 328 K. The volume of each gas in a mixture is equal to the volume of the container, so V = 6.0 L.
PNe

nNe
RTV

(0.12 mol)(0.0821 L
⋅atm
⋅K
-1⋅
mol
-1)(328 K)
6.0 L
= 0.54 atm
PHe

nHe
RTV

(0.19 mol)(0.0821 L
⋅atm
⋅K
-1⋅
mol
-1)(328 K)
6.0 L
= 0.85 atm
The total pressure is the sum of the partial pressures: P
total
= 0.54 + 0.85 = 1.39 atm.
Alternatively, the total pressure can be obtained by applying the ideal gas law directly to the mixture. n
total
= 0.12 mol Ne + 0.19 mol He = 0.31 mol.
Ptotal

ntotal
RTV

(0.31 mol)(0.0821 L
⋅atm
⋅K
-1⋅
mol
-1)(328 K)
6.0 L
= 1.39 atm
Chapter 7 States of Matter and Changes in State
© by
North
Carolina
State
University