Question

At ${127}^{0}C$ and 1 atm pressure, a mixture of a gas contains 0.3 mole of $N_2$ and 0.2 mole of $O_2$. The volume of the mixture is?

• A. 15 L
• B. 18.2 L
• C. 16.4 L
• D. 22.4 L

Answer 1

The correct option is C 16.4 L

Let $V$ be volume of mixture.

Partial pressure of $O_2=X_{O_2}\times P_{total}=\frac{n_{O_2}RT}{V}$

$\therefore V=\frac{n_{O_2}RT}{X_{O_2}\times P_{total}}=\frac{0.2\times 0.0821\times 400}{\left(\frac{0.2}{0.5}\right)\times 1}=16.4L$