| ⇦ |
| ⇨ |
Assuming fully decomposed, the volume of CO₂ released at STP on heating 9.85 g of BaCO₃(Atomic mass Ba=137) will be
Options
(a) 2.24 L
(b) 4.96 L
(c) 1.12L
(d) 0.84L
Correct Answer:
1.12L
Explanation:
BaCO₃ → BaO+ CO₂
Atomic Mass of Ba=137
Atomic Mass of C=12
Atomic Mass of O=16
Molecular Mass of BaCO₃ = 137 + 12 + (16*3) = 197
197 gm of BaCO₃ released carbondioxide = 22.4 litre at STP
1 gm of BaCO₃ released carbondioxide = 22.4/197 litre
9.85 gm of BaCO₃ released carbondioxide = 22.4/197 x 9.85 = 1.12 litre
Related Questions: - The neutralisation of a strong acid by a strong base liberates an amount
- In the manufacture of ethanol from starch by fermentation,
- Treatment of acetaldehyde with ethyl magnesium bromide and subsequent hydrolysi
- On heating sodium metal in dry ammonia, the compound formed is
- In which of the solutions containing following solutes, its normality is equal
Question Type: Memory
(964)
Difficulty Level: Easy
(1008)
Topics: Basic Concepts of Chemistry
(94)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The neutralisation of a strong acid by a strong base liberates an amount
- In the manufacture of ethanol from starch by fermentation,
- Treatment of acetaldehyde with ethyl magnesium bromide and subsequent hydrolysi
- On heating sodium metal in dry ammonia, the compound formed is
- In which of the solutions containing following solutes, its normality is equal
Question Type: Memory (964)
Difficulty Level: Easy (1008)
Topics: Basic Concepts of Chemistry (94)
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