| ⇦ | 
| ⇨ |
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: - The pH value of a 10 M solution of HCl is
- Of the following, which change will shift the reaction towards the product
- Mg burns in CO₂ to produce
- Number of unpaired electrons in N2⁺ is
- The basicity of aniline is weaker in comparison to that of mathyl amine due to
Topics: Solid State and Solutions
(91)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The pH value of a 10 M solution of HCl is
- Of the following, which change will shift the reaction towards the product
- Mg burns in CO₂ to produce
- Number of unpaired electrons in N2⁺ is
- The basicity of aniline is weaker in comparison to that of mathyl amine due to
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