The molar 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

Correct Answer:

0.34

Explanation:

Given: Weight of nitrogen = 70g; Weight of oxygen = 120 g and weight of carbon dioxide = 44 grams.

Moles of N₂(n₁) = Weight/ Molecular weight = 70/28 = 2.5.

Similarly, moles of O₂(n₂) = 120/32 = 3.75, and moles of CO₂(n₃) = 44/44 = 1.

Therefore mole fraction of nitrogen(N₂) = n₁/ n₁+n₂+n₃ = 2.5/ 2.5+3.75+1 =2.5/7.25 =0.34.