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Test: Atoms and Molecules- Case Based Type Questions - BMAT MCQ


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15 Questions MCQ Test - Test: Atoms and Molecules- Case Based Type Questions

Test: Atoms and Molecules- Case Based Type Questions for BMAT 2024 is part of BMAT preparation. The Test: Atoms and Molecules- Case Based Type Questions questions and answers have been prepared according to the BMAT exam syllabus.The Test: Atoms and Molecules- Case Based Type Questions MCQs are made for BMAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Atoms and Molecules- Case Based Type Questions below.
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Test: Atoms and Molecules- Case Based Type Questions - Question 1
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Direction: The following data represents the distribution of electrons, protons and neutrons in atoms of four elements A, B, C, D. Understand the data carefully and answer the following questions.

Q. Select the correct electronic distribution of element B:

Detailed Solution for Test: Atoms and Molecules- Case Based Type Questions - Question 1
The electronic configuration of element B = 2, 8, 6
Test: Atoms and Molecules- Case Based Type Questions - Question 2
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Direction: The following data represents the distribution of electrons, protons and neutrons in atoms of four elements A, B, C, D. Understand the data carefully and answer the following questions.

Q. The Mass number of element D is:

Detailed Solution for Test: Atoms and Molecules- Case Based Type Questions - Question 2
Mass number = Number of Protons + Number of Neutrons

= 17 + 22 = 39

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Test: Atoms and Molecules- Case Based Type Questions - Question 3
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Direction: The following data represents the distribution of electrons, protons and neutrons in atoms of four elements A, B, C, D. Understand the data carefully and answer the following questions.

Q. The atomic number of element B is:

Detailed Solution for Test: Atoms and Molecules- Case Based Type Questions - Question 3
Atomic number of element = Number of Protons in that element = Number of Electrons.
Test: Atoms and Molecules- Case Based Type Questions - Question 4
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Direction: The following data represents the distribution of electrons, protons and neutrons in atoms of four elements A, B, C, D. Understand the data carefully and answer the following questions.

Q. The Valency of element A:
Detailed Solution for Test: Atoms and Molecules- Case Based Type Questions - Question 4
The electronic configuration of element A is 2, 7. There are seven valence electrons in its outermost shell.

So, valency = (8 − no. of valence electrons)

= 8 − 7 = 1

Test: Atoms and Molecules- Case Based Type Questions - Question 5
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Direction: Sanjana observed that when 3.0 gm of carbon is burnt in 8.0 gm of oxygen, 11.0 gm of carbon dioxide is produced. Based on the given information, answer the following questions.

Q. Name the Law of Chemical Combination shown in the above passage:
Detailed Solution for Test: Atoms and Molecules- Case Based Type Questions - Question 5
According to this law, the elements are always present in definite proportion by mass in a chemical substance. All pure samples of a compound contain the same elements combined together in the same proportion by mass.
Test: Atoms and Molecules- Case Based Type Questions - Question 6
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Direction: Sanjana observed that when 3.0 gm of carbon is burnt in 8.0 gm of oxygen, 11.0 gm of carbon dioxide is produced. Based on the given information, answer the following questions.

Q. In what ratio does carbon and oxygen combine to form carbon dioxide?
Detailed Solution for Test: Atoms and Molecules- Case Based Type Questions - Question 6
When 3.0 gm of carbon is burnt in 8.0 gm oxygen, 11.0 gm of carbon dioxide is produced. It means carbon and oxygen are combined in the ratio of 3 : 8 to form carbon dioxide. Thus, when there is 3 gm carbon and 50 gm oxygen, then also only 8 gm of oxygen will be used and 11 gm of carbon dioxide will be formed. The remaining oxygen is not used.
Test: Atoms and Molecules- Case Based Type Questions - Question 7
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Direction: Sanjana observed that when 3.0 gm of carbon is burnt in 8.0 gm of oxygen, 11.0 gm of carbon dioxide is produced. Based on the given information, answer the following questions.

Q. In a compound water at what ratio hydrogen and oxygen combine to form water:
Detailed Solution for Test: Atoms and Molecules- Case Based Type Questions - Question 7
The molecular formula of water is H2O and its molecular mass is 18g. So, when 2 g of Hydrogen combines with 16 g of Oxygen then 18 g of water is formed. So, the ratio is

H : O

2 : 16

1 : 8

Test: Atoms and Molecules- Case Based Type Questions - Question 8
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Direction: Sanjana observed that when 3.0 gm of carbon is burnt in 8.0 gm of oxygen, 11.0 gm of carbon dioxide is produced. Based on the given information, answer the following questions.

Q. In a chemical substance, elements are present in a definite proportion by ______.
Detailed Solution for Test: Atoms and Molecules- Case Based Type Questions - Question 8
The elements are present in a definite proportion by mass.
Test: Atoms and Molecules- Case Based Type Questions - Question 9
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Direction: Two class students of class 9th, Aashi and Sheena, were asked to take 5.3 g of sodium carbonate and 6 g of ethanoic acid to make 2.2 g of carbon dioxide, 0.9 g of water and 8.2 g of sodium ethanoate. Aashi followed the instructions but Sheena took the chemicals without measuring their amounts.

Q. Which law does this agreements shows?
Detailed Solution for Test: Atoms and Molecules- Case Based Type Questions - Question 9
5.3 g + 6 g → 2.2g + 0.9 g + 8.2 g 11.3 g → 11.3 g

This agreement shows the “Law of Conservation of mass”.

Test: Atoms and Molecules- Case Based Type Questions - Question 10
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Direction: Two class students of class 9th, Aashi and Sheena, were asked to take 5.3 g of sodium carbonate and 6 g of ethanoic acid to make 2.2 g of carbon dioxide, 0.9 g of water and 8.2 g of sodium ethanoate. Aashi followed the instructions but Sheena took the chemicals without measuring their amounts.

Q. Whose activity do you think will be in agreement with the law?
Detailed Solution for Test: Atoms and Molecules- Case Based Type Questions - Question 10
Aashi activity will be in agreement with the law.
Test: Atoms and Molecules- Case Based Type Questions - Question 11
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Direction: Two class students of class 9th, Aashi and Sheena, were asked to take 5.3 g of sodium carbonate and 6 g of ethanoic acid to make 2.2 g of carbon dioxide, 0.9 g of water and 8.2 g of sodium ethanoate. Aashi followed the instructions but Sheena took the chemicals without measuring their amounts.

Q. The Law states that _______ can neither be created nor be destroyed in a chemical reaction.
Detailed Solution for Test: Atoms and Molecules- Case Based Type Questions - Question 11

  • The law states that mass can neither be created nor be destroyed.

  • The Law of Conservation of Mass dates from Antoine Lavoisier's 1789 discovery that mass is neither created nor destroyed in chemical reactions.

  • In other words, the mass of any one element at the beginning of a reaction will equal the mass of that element at the end of the reaction.

Test: Atoms and Molecules- Case Based Type Questions - Question 12
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Direction: Rahul took 5 moles of carbon atoms in a container and Sohan also took 5 moles of sodium atoms in another container of same weight.

Q. Whose container has more number of atoms?
Detailed Solution for Test: Atoms and Molecules- Case Based Type Questions - Question 12
Both containers have same number of atoms since they contain same number of moles.
Test: Atoms and Molecules- Case Based Type Questions - Question 13
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Direction: Rahul took 5 moles of carbon atoms in a container and Sohan also took 5 moles of sodium atoms in another container of same weight.

Q. Which container is heavier?
Detailed Solution for Test: Atoms and Molecules- Case Based Type Questions - Question 13
Mass of container containing 5 moles of C atoms = 5 × 12 = 60 g

Mass of container containing 5 moles of Na atoms = 5 x 23 = 115 g

Hence, container containing 5 moles of sodium is heavier.

Test: Atoms and Molecules- Case Based Type Questions - Question 14
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Direction: Rahul took 5 moles of carbon atoms in a container and Sohan also took 5 moles of sodium atoms in another container of same weight.

Q. Number of atoms in one mole:
Detailed Solution for Test: Atoms and Molecules- Case Based Type Questions - Question 14
1 mole = 6.022 × 1023 atoms.
Test: Atoms and Molecules- Case Based Type Questions - Question 15
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Direction: Rahul took 5 moles of carbon atoms in a container and Sohan also took 5 moles of sodium atoms in another container of same weight.

Q. The exact number of atoms present in 12 gm of Carbon-12:
Detailed Solution for Test: Atoms and Molecules- Case Based Type Questions - Question 15
The exact number of atoms present in 12 gm of Carbon-12 is called Avogadro’s constant.
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