| ⇦ | 
| ⇨ |
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: - 6.02 x 10²° molecules of urea are present in 100 mL of its solution
- Molecule in which the distance between the two adjacent carbon atoms is largest
- In the formation of N₂⁺from N₂, the electron is lost from
- In liquid-gas equilibrium,the pressure of vapours above the liquid is constant
- The compound benzophenone(C₆H₅COC₆H₅) does not reacts with
Topics: Solid State and Solutions
(91)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- 6.02 x 10²° molecules of urea are present in 100 mL of its solution
- Molecule in which the distance between the two adjacent carbon atoms is largest
- In the formation of N₂⁺from N₂, the electron is lost from
- In liquid-gas equilibrium,the pressure of vapours above the liquid is constant
- The compound benzophenone(C₆H₅COC₆H₅) does not reacts with
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