Problem: A gas mixture contains 1.26 g N2 and 0.86 g O2 in a 1.56-L container at 18 oC.You want to reference (Pages 216 - 222) Section 5.6 while completing this problem.Calculate the partial pressure of O2.

We are asked to calculate the partial pressure of O₂.

Molar mass O₂ = 32.00 g/mol

Molar mass N₂ = 28.02 g/mol

Calculate moles each: 

$$\begin{gathered} \mathbf{mol}\mathbf{ }{\mathbf{O}}_{\mathbf{2}}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{1}\mathbf{.}\mathbf{26}\cancel{\mathbf{ }\mathbf{g}\mathbf{ }{\mathbf{O}}_{\mathbf{2}}\mathbf{ }}\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\mathbf{mol}\mathbf{ }\mathbf{ }{\mathbf{O}}_{\mathbf{2}}}{\mathbf{32}\mathbf{.}\mathbf{00}\cancel{\mathbf{ }\mathbf{g}\mathbf{ }{\mathbf{O}}_{\mathbf{2}}\mathbf{ }}}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{0}\mathbf{.}\mathbf{039}\mathbf{ }\mathbf{mol}\mathbf{ }{\mathbf{O}}_{\mathbf{2}}\mathbf{ } \\ \mathbf{mol}\mathbf{ }{\mathbf{N}}_{\mathbf{2}}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{0}\mathbf{.}\mathbf{86}\cancel{\mathbf{ }\mathbf{g}\mathbf{ }{\mathbf{N}}_{\mathbf{2}}\mathbf{ }}\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\mathbf{mol}\mathbf{ }\mathbf{ }{\mathbf{N}}_{\mathbf{2}}}{\mathbf{28}\mathbf{.}\mathbf{02}\cancel{\mathbf{ }\mathbf{g}\mathbf{ }{\mathbf{N}}_{\mathbf{2}}\mathbf{ }}}\mathbf{=}\mathbf{ }\mathbf{0}\mathbf{.}\mathbf{031}\mathbf{ }\mathbf{mol}\mathbf{ }{\mathbf{N}}_{\mathbf{2}} \end{gathered}$$

Calculate mole fraction O₂ :

$$\begin{gathered} \boxed{{\mathbf{X}}_{{\mathbf{O}}_{\mathbf{2}}}\mathbf{ }\mathbf{=}\mathbf{ }\frac{\mathbf{mol}\mathbf{ }{\mathbf{O}}_{\mathbf{2}}}{\mathbf{mol}\mathbf{ }\mathbf{total}}} \\ {\mathbf{X}}_{{\mathbf{O}}_{\mathbf{2}}}\mathbf{ }\mathbf{=}\mathbf{ }\frac{\mathbf{mol}\mathbf{ }{\mathbf{O}}_{\mathbf{2}}}{\mathbf{mol}\mathbf{ }\mathbf{total}} \\ {\mathbf{X}}_{{\mathbf{O}}_{\mathbf{2}}}\mathbf{ }\mathbf{=}\mathbf{ }\frac{\mathbf{0}\mathbf{.}\mathbf{039}\mathbf{ }\mathbf{mol}}{\mathbf{0}\mathbf{.}\mathbf{039}\mathbf{ }\mathbf{mol}\mathbf{ }\mathbf{+}\mathbf{ }\mathbf{0}\mathbf{.}\mathbf{031}\mathbf{ }\mathbf{mol}} \\ {\mathbf{X}}_{{\mathbf{O}}_{\mathbf{2}}}\mathbf{ }\mathbf{=}\mathbf{ }\frac{\mathbf{0}\mathbf{.}\mathbf{039}\mathbf{ }\mathbf{mol}}{\mathbf{0}\mathbf{.}\mathbf{070}\mathbf{ }\mathbf{mol}} \end{gathered}$$

X_O2 = 0.557 

Given:

R = 0.08206 L-atm/mol-K

T = 18 ^oC +273.15 = 291.15 K

V= 1.56 L

n = 0.070 mol

Calculate total pressure: 

$$\begin{gathered} \boxed{\mathbf{PV}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{nRT}} \\ \mathbf{P}\mathbf{ }\mathbf{=}\frac{\mathbf{ }\mathbf{nRT}}{\mathbf{V}} \\ \mathbf{P}\mathbf{ }\mathbf{=}\frac{\mathbf{ }\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{070}\mathbf{ }\cancel{\mathbf{mol}}\mathbf{)}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{08206}\mathbf{ }{\displaystyle \frac{ \cancel{L}-\mathrm{atm}}{ \cancel{\mathrm{mol}}- \cancel{K}}}\mathbf{)}\mathbf{(}\mathbf{291}\mathbf{.}\mathbf{15}\mathbf{ }\cancel{\mathbf{K}}\mathbf{)}}{\mathbf{1}\mathbf{.}\mathbf{56}\mathbf{ }\cancel{\mathbf{L}}} \end{gathered}$$

P = 1.07 atm