Question

A neon-dioxygen mixture contains $70.6\text{g}$ dioxygen and $167.5\text{g}$ 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?

• A. $P_{O_2}=19.75\text{bar},P_{Ne}=5.25\text{bar}$
• B. $P_{O_2}=12.75\text{bar},P_{Ne}=12.25\text{bar}$
• C. $P_{Ne}=19.75\text{bar},P_{O_2}=5.25\text{bar}$
• D. $P_{Ne}=12.75\text{bar},P_{O_2}=12.25\text{bar}$

Answer 1

The correct option is C $P_{Ne}=19.75\text{bar},P_{O_2}=5.25\text{bar}$

Given :
Mass of dioxygen $=70.6\text{g}$
Mass of Neon $=167.5\text{g}$
Total pressure $=25\text{bar}$
Number of moles of dioxygen $=\frac{70.6\text{g}}{32\text{g}}=2.21\text{mol}$
Number of moles of neon $=\frac{167.5\text{g}}{20\text{g}}=8.375\text{mol}$
Mole fraction of dioxygen $=\frac{2.21}{2.21+8.375}=0.21$
$\text{Mole fraction of neon}=1-0.21=0.79$
Partial pressure of gas = mole fraction $\times$ total pressure
$\Rightarrow$ Partial pressure of a dioxygen= $0.21\times \left(25\text{bar}\right)=5.25\text{bar}$
Partial pressure of neon $=0.79\times \left(25\text{bar}\right)=19.75\text{bar}$
Hence, the correct answer is option (c).