Test: Atoms and Molecules- Case Based Type Questions


15 Questions MCQ Test Online MCQ Tests for Class 9 | Test: Atoms and Molecules- Case Based Type Questions


Description
Attempt Test: Atoms and Molecules- Case Based Type Questions | 15 questions in 30 minutes | Mock test for Class 9 preparation | Free important questions MCQ to study Online MCQ Tests for Class 9 for Class 9 Exam | Download free PDF with solutions
QUESTION: 1

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:

Solution: The electronic configuration of element B = 2, 8, 6
QUESTION: 2

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:

Solution: Mass number = Number of Protons + Number of Neutrons

= 17 + 22 = 39

QUESTION: 3

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:

Solution: Atomic number of element = Number of Protons in that element = Number of Electrons.
QUESTION: 4

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:

Solution: 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

QUESTION: 5

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:

Solution: 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.
QUESTION: 6

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?

Solution: 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.
QUESTION: 7

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:

Solution: 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

QUESTION: 8

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 ______.

Solution: The elements are present in a definite proportion by mass.
QUESTION: 9

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?

Solution: 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”.

QUESTION: 10

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?

Solution: Aashi activity will be in agreement with the law.
QUESTION: 11

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.

Solution:
  • 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.

QUESTION: 12

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?

Solution: Both containers have same number of atoms since they contain same number of moles.
QUESTION: 13

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?

Solution: 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.

QUESTION: 14

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:

Solution: 1 mole = 6.022 × 1023 atoms.
QUESTION: 15

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:

Solution: The exact number of atoms present in 12 gm of Carbon-12 is called Avogadro’s constant.
Use Code STAYHOME200 and get INR 200 additional OFF
Use Coupon Code