Question

The mole fraction of nitrogen, in a mixture containing 70 grams nitrogen, 120 grams of oxygen and 44 grams of carbon dioxide is

• A. 0.36
• B. 0.34
• C. 0.29
• D. 5.0

Answer 1

Answer: (B)

Weight of nitrogen $=70g$;

Weight of oxygen $=120g$

and weight of carbon dioxide $=44g$.

Moles of $N_2\left(n_1\right)=\frac{\text{ Weight }}{\text{ Molecular weight }}$

$=\frac{70}{28}=2.5$

Similarly, moles of $O_2\left(n_2\right)=\frac{120}{32}=3.75$

and moles of $C{O}_2\left(n_3\right)=\frac{44}{44}=1$

Therefore mole fraction of nitrogen $\left(N_2\right)$

$=\frac{n_1}{n_1+n_2+n_3}$

$=\frac{2.5}{2.5+3.75+1}$

$=\frac{2.5}{7.25}=0.34$