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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: - Benzaldehyde + NaOH produces which of the following
- Which of the following 0.10 m aq solution will have the lowest freezing point
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Topics: Solid State and Solutions
(91)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Benzaldehyde + NaOH produces which of the following
- Which of the following 0.10 m aq solution will have the lowest freezing point
- At 25⁰ C , the highest osmotic pressure is exhibited by 0.1 M solution of
- An open system
- For the reaction C(s) + O₂(g)
Topics: Solid State and Solutions (91)
Subject: Chemistry (2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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