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

Options

A. 0.36
B. 0.34
C. 0.29
D. 5

Answer

B

Explanation

Weight of nitrogen = 70 g, Weight of oxygen = 120 g and weight of carbon dioxide = 44 g.
Moles of `N_2 (n_1) = ("Weight")/("Molecular weight")= 70/28 =2.5`
Similarly, moles of `O_2 (n_2) = 120/32 = 3.75`
and moles of `CO_2 (n_3) = 44/44= 1`
Therefore mole fraction of nitrogen `(N_2)`
`= (n_1)/(n_1+n_2 + n_3) = (2.5)/(2.5+3.75 +1)= (2.5)/(7.25)=0.34`