| ⇦ |
| ⇨ |
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.
Related Questions: - Which of the esters shown, after reduction with LiAlH₄ and aqueous workup, will
- How much copper is supposed to be deposited when a current of 0.75 amperes
- Heat of formation of H₂O(g) at 25⁰C is -243 kJ,ΔE for the reaction
- Diffusion of solvent through a semi permeable membrane is called
- Least voltage hydrogen halide is
Topics: Solid State and Solutions
(91)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Which of the esters shown, after reduction with LiAlH₄ and aqueous workup, will
- How much copper is supposed to be deposited when a current of 0.75 amperes
- Heat of formation of H₂O(g) at 25⁰C is -243 kJ,ΔE for the reaction
- Diffusion of solvent through a semi permeable membrane is called
- Least voltage hydrogen halide is
Topics: Solid State and Solutions (91)
Subject: Chemistry (2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends

Leave a Reply