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
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