Question

A mixture of nitrogen and oxygen has density of $1.00gL$, and molecular weight (Mav) $31.2g$ at ${27}^{o}C$ and $600mmHg$. What is mole fraction of $N_2$ ?

• A. $0.2$
• B. $0.4$
• C. $0.6$
• D. $0.8$

Answer 1

The correct option is A $0.2$

$\begin{array}{c}\text{Given: -}\\ \text{Mavg}=31.2g\text{at}{27}^{\circ }C\text{and}600mm\text{of Hg}\\ \text{for mixture of}{N}_2\text{and}{O}_2(\rho =\lg /L)\end{array}$

we know:-

$\begin{array}{cc}\text{Mavg}& =X_{N_2}\times M_{N_2}+X_{O2}\times M_{O2}\\ & =28X_{N2}+32X_{O2}\end{array}$

$31\cdot 2=28X_{N2}+32X_{O2}-\left(1\right)$

also $X_{N2}+X_{O2}=1-\left(2\right)$

putting (2) in (1)

$31\cdot 2=28(1-X{o}_2)+32X{o}_2$

$31\cdot 2-28=4X{o}_2$

$\Rightarrow X{o}_2=0.8$ and $X_{N2}=0.2$

So, $A$ ) 0.2 is correct answer