question_answer 1)

A neon gas-dioxygen gas mixture contains 70.6 g dioxygen and 167.5 g neon. If the pressure of the mixture of gases in the cylinder is 25 bar, what is the partial pressure of dioxygen and neon in the mixture?

Answer:

Information shadow: Mass of neon = 167.5 g ; Molar mass of neon = \[20g\,mo{{l}^{-1}}\] Mass of dioxygen = 70.6 g Molar mass of dioxygen = \[32g\,mo{{l}^{-1}}\] Total pressure of the mixture = 25 bar Problem solving strategy: Partial pressure of a gas = Mole fraction of the gas \[\times \]Total pressure of the mixture Working it out: \[{{n}_{{{O}_{2}}}}=\frac{70.6}{32}=2.21\,\,;\,\,{{n}_{Ne}}=\frac{167.5}{20}=8.375\] \[{{x}_{{{O}_{2}}}}=\frac{{{n}_{{{O}_{2}}}}}{{{n}_{{{O}_{2}}}}+{{n}_{Ne}}}=\frac{2.21}{2.21+8.375}=0.21\] \[{{x}_{Ne}}=1-{{x}_{{{O}_{2}}}}=1-0.21=0.79\] \[{{P}_{{{O}_{2}}}}={{x}_{{{O}_{2}}}}\times P\] \[=0.21\times 25\] \[=5.25bar\] \[{{P}_{Ne}}={{x}_{Ne}}\times P\] \[=0.79\times 25\] \[=19.75\]bar