MSBSHSE Class 9 Science Chapter 4 Measurement of Matter Solutions

MSBSHSE Class 9 Science Chapter 4 Measurement of Matter Solutions is used by students to prepare most competently for the exams. It is a very significant resource from the standpoint of the Science examination. The solutions come with detailed answers and step by step explanations, covering the questions covered in the chapter from the MSBSHSE Class 9 Science textbook. These solutions will help the students to understand all the basic concepts of the chapter.

The crucial topic, Measurement of Matter from Science of Class 9 is discussed in depth in this chapter and students can master the concepts by referring to these MSBSHSE Class 9 Solutions of Science Chapter 4. Laws of chemical combination, Atom – shape, mass, valency, Radicals and so on are topics covered in this chapter. These solutions are prepared after proper research and will help the students to get a very thorough conceptual understanding. The well-structured content makes it easier for the students to learn the concepts. The content is as per the latest MSBSHSE Syllabus for Class 9 and the solutions provided here are designed to help ace the exams. So, solving these questions with highly relevant answers provides help to the students to master the subject and set the foundation for higher classes.

Maharashtra Board Class 9 Science Chapter 4- BYJU’S Important Questions & Answers

1. Give examples.

a. Positive radicals

b. Basic radicals

c. Composite radicals

d. Metals with variable valency

e. Bivalent acidic radicals

f. Trivalent basic radicals

Answer: a. Positively charged ions are known as cations. Some examples of positive radicals include H ⁺ Hydrogen and Na⁺ Sodium

b. Cation is known as basic radicals. Hence, H ⁺ Hydrogen, Na⁺ Sodium, Cu²⁺ Copper are all examples of Basic radicals

c. Composite radicals is a radical containing a group of atoms carrying a charge. For example, Acetylimino, Cyclohexane carbonyl and so on.

d. In the scenarios, when the atoms of some elements are given away or have taken up different numbers of electrons, those elements exhibit more than one valency. This is known as variable valency. Examples of variable valency are Copper(Cu) that exhibits the variable valencies 1 and 2: Cu⁺ and Cu^{2 +} and Mercury(Hg) that exhibit the variable valencies 1 and 2: Hg⁺ and Hg^{2 +}_.

e. An element whose atom can replace 2 atoms of hydrogen or another univalent element is known as Bivalent acidic radicals and is also referred as a radical that has the same valence as a bivalent atom. (SO₄) ²– and (CO₃) ^2- are examples of bivalent acidic radicals.

f. A cation that has a valency of three is a trivalent basic radical. Examples of trivalence basic radicals are Al^{3 +} ,Cr^{3 +} , Bi^{3 +} and Fe ^{+ 2}_.

2. Write symbols of the following elements and the radicals obtained from them, and indicate the charge on the radicals.

(Mercury, potassium, nitrogen, copper, sulphur, carbon, chlorine, oxygen)

Answer: Radicals are the unpaired valence electrons on an atom, ion or molecule.

3. Write the steps in deducing the chemical formulae of the following compounds.

(Sodium sulphate, potassium nitrate, ferric phosphate, calcium oxide, aluminium hydroxide)

Answer: Chemical formula is created on the basis of the cation and anion found in the compound.

Sodium sulphate– Step 1 : Give the symbols of the radicals with basic radical on the left

Na SO₄

Step 2: Write the valency below the respective radical

Na SO₄

1 2

Step 3- Cross multiply the number of radicals as shown with the arrow

Step 4: Thus write down the chemical formula of the compound

Na₂SO₄

Use the method for all the other compounds.

Potassium Nitrate– Step 1- K NO₃

Step 2- K NO₃

1 1

And so chemical formula is KNO₃

Ferric Phosphate – Fe PO₄

Fe PO₄

3 3

Hence, chemical formula is FePO₄

Calcium Oxide– Basic radical on the left

Ca O

Cross multiply the radicals with the valence mentioned below the respective radical

Ca O

3 3

Thus, the chemical formula is CaO.

Aluminium hydroxide- Write the symbols with the basic radical to the left

Al OH

Now, cross multiply with the valences listed under the respective radical.

Al OH

3 1

Thus, the chemical formula of the compound is Al(OH)₃.

4. Explain how the element sodium is monovalent.

Answer: Sodium is an element with atomic number 11 and has one valence electron in its valence or external shell, which needs to be removed so as to attain the noble gas configuration. Hence, it gets ionized and loses one electron to another compound to form a compound. Hence, the valency of the atom is 1 and so is called a monovalent atom.

5. M is a bivalent metal. Write down the steps to find the chemical formulae of its compounds formed with the radicals, sulphate and phosphate.

Answer: M is a bivalent metal as it has valency 2 (M²⁺ ). Given below are the steps to find the chemical formula of its compounds formed with the radicals, sulphate and phosphate:

First, take metal to form a compound with sulfate. Hence,

M SO₄

Now write the valency of both components (metal is 2 as it is bivalent and sulphur is 2 as it belongs to the halogen family). Now, cross multiplying the radicals with the valencies you get

MSO_4, as one ion of metal M and one sulphate ion will balance each other.

Phosphate(PO₄)^3-, in the meantime, is a trivalent anion. So, given 2 valence of metals with 3 valence of phosphate to cross multiply, you get the chemical formula M₃(PO₄)₂.

6. Explain the need for a reference atom for atomic mass. Give some information about two reference atoms.

Answer: Mass of an atom, concentrated at its nucleus is the sum of proton(p) and neutron). The number (p+n) is called the atomic mass number. Now, since we know that the atom is small in size, how do we determine its mass? Take into consideration the concept of reference atom for atomic mass. Hydrogen was selected as the reference atom because it was the lightest. The relative mass of hydrogen was 1, i.e, the relative mass of nitrogen was calculated as 14 times more than hydrogen. Later on Carbon came to be accepted as the reference atom. Now, the relative mass of carbon was 12. Thus, the relative atomic mass of hydrogen as to the carbon atom is 12 × 1/12= 1.

7. What is meant by Unified Atomic Mass ?

Answer: A standard unit of mass that quantifies mass on an atomic molecular scale is called Unified Atomic Mass unit or Dalton. Its symbol is ‘u’. 1u = 1.66053904 × 10^-27 kg.

8. Explain with examples what is meant by a ‘mole’ of a substance.

Answer: A mole of a compound is that quantity of a substance whose mass in grams is equal in magnitude to its molecular mass in Daltons. The SI unit is mol. Thus, the molecular mass of oxygen is 32u, and therefore 32g oxygen is 1 mole of oxygen. The molecular mass of water is 18u. Therefore, 18g of water makes 1 mole of water.

9. Write the names of the following compounds and deduce their molecular masses.

(Na₂SO₄, K₂CO₃, CO₂, MgCl₂, NaOH, AlPO₄, NaHCO₃).

Answer:

Na₂SO₄ – Molecular mass of Sodium Sulfate = 2(Atomic mass of Na) + 1(Atomic mass of S) + 4(Atomic mass of O) = 2 × 23 + 1× 32 + 4× 16

= 46 + 32 + 64 = 142 g/mol.

K₂CO₃– Molecular mass of Potassium Carbonate

₌ 2 × 39 + 1× 12 + 3× 16= 78 + 12 + 48

= 138 g/mol

CO_{2 –} Molecular mass of Carbon Dioxide

= 1× 12 + 2× 16= 12 + 32

= 44 g/mol

MgCl₂– Molecular mass of Magnesium Chloride

= 1× 24 + 2× 35.5 = 24 + 70

= 94 g/mol

NaOH- Molecular mass of Sodium Hydroxide

= 1× 23 + 1× 16 + 1× 1

= 23 + 16 + 1

= 40 g/mol

AlPO₄– Molecular mass of Aluminium Phosphate

=1× 27 + 1× 31 + 4× 16

= 27 + 31 + 64

= 122 g/mol

NaHCO3- Molecular mass of Sodium Bicarbonate

= 1× 23 + 1× 1 + 1× 12 + 3× 16 = 23 + 1 + 12 + 48

= 84 g/mol

10. Two samples ‘m’ and ‘n’ of slaked lime were obtained from two different reactions. The details about their composition are as follows:

‘sample m’ mass : 7g

Mass of constituent oxygen : 2g

Mass of constituent calcium : 5g

‘sample n’ mass : 1.4g

Mass of constituent oxygen : 0.4g

Mass of constituent calcium : 1.0g

Which law of chemical combination does this prove? Explain.

Answer:

Given that,

‘sample m’ mass : 7g

Mass of constituent oxygen : 2g

Mass of constituent calcium : 5g

‘sample n’ mass : 1.4g

Mass of constituent oxygen : 0.4g

Mass of constituent calcium : 1.0g

Of all the laws of combination,“ Law of Constant Proportion” proves this.

As per the Law of Constant Proportion, The proportion by weight of the constituent elements in the various samples of a compound is fixed in a ratio.” Take the example of the proportion by weight of hydrogen and oxygen in water is 1:8.

Now, from the given data, how do we prove it?

Now, take the sample of “m” , the ratio of proportion of calcium and oxygen is 5:2

Likewise, if we take sample “n”, the ratio of proportion of calcium and oxygen is 1.0: 0.4

Therefore, Ca: 0 = 1.0: 0.4 = 10: 4 =5:2.

Hence, by simplifying the ratio proportion by mass you can get the same values that verify “The Law of Constant Proportion”.

11. Deduce the number of molecules of the following compounds in the given quantities.

(32g oxygen, 90g water, 8.8g carbon dioxide, 7.1g chlorine).

Answer: 32 g Oxygen

Find Number of moles in O₂ (n) = Mass of substance in grams / molecular mass of a substance

Therefore n = 32 /32 =1 mole

There are 6.022 x 10²³ molecules in 1 mol.

Now, take the example of 90g water

Number of moles in water = Mass of water in grams / molecular mass = 90 / 18 = 5 moles

Hence, no of molecules = 6.022 x 10²³x 5 = 30.11 x 10²³ molecules

Take 8.8 g carbon dioxide

Number of moles in water = Mass of water in grams / molecular mass= 8.8/40= 0.2 mol

Therefore, no of molecules = 0.2 x 6.022 x 10²³= 1.2044 x 10²³ molecules

7.1 g chlorine

Number of moles in water = Mass of water in grams / molecular mass = 7.1/71= 0.1 mol

Thus, no of molecules = 0.1 x 6.022 x 10²³= 0.6022 x 10²³.

12. If 0.2 mol of the following substance are required how many grams of those substances should be taken?

(Sodium chloride, magnesium oxide, calcium carbonate)

Answer: No of moles of substance= mass of the substance/ molecular mass of the substance

Hence, molar mass = sum of constituent atomic masses.

Thus Molar Mass of Sodium Chloride (NaCl) = 23 + 35.5 = 58.5 g/ mol

Given that number of moles of a substance= mass of the substance/ molecular mass of the substance

0.2 = x/ 11.7 g

Hence, it is seen that you need 11.7 g of NaCl to obtain 0.2 mol Sodium Chloride.

Now, consider Magnesium Oxide (MgO) = 24 +16= 40 g/ mol

Since, number of moles of a substance= mass of the substance/ molecular mass of the substance

0.2 = x / 40 = 8g

So, 8g of Mg0 is required to obtain 0.2 mol Magnesium Oxide

Then, take the example of CaCO₃.

Molar mass of CaCO₃= 40 +12 + 3 x 16

= 100 g/ mol

Number of moles of a substance= mass of the substance/ molecular mass of the substance

= mass of the substance in grass /

0.2 = x / 200= 20g

Hence, 20g CaCO₃ is required to get 0.2 moles of CaCO_{3 .}

13. Name the scientists who discovered the laws of chemical combination.

Answer: Dalton, Thomson and Rutherford, the scientists studied the structure of substances and the atom and thus discovered the laws of chemical combination.

14. What did the French Scientist Antoine Lavoisier infer from his research in 1785?

Answer: The French Scientist, Antoine Lavoisier inferred from his research in 1785 that ‘there is no rise or drop in the weight of the matter during a chemical reaction.’

15. How is the size of an atom determined?

Answer: The radius of an atom determines its size. Atomic radius is expressed in nanometers. Hence, the atomic radius of an isolated atom is the distance between the nucleus of an atom and its outermost orbit. Meanwhile, the atomic size is also dependent on the number of electron orbits in the atom. If the number of orbits is greater than the atomic size is larger.

16. What are nucleons?

Answer: Protons and neutrons together are called nucleons.

17. Which are the atoms that have independent existence? What does it mean?

Answer: Some of the atoms of elements such as helium, neon exist independently. It means that these elements are in a mono-atomic molecular state.

19. What is a polyatomic molecular state?

Answer: At times, two or more atoms of an element combine to produce molecules of that element. Such elements are in a polyatomic molecular state. The elements oxygen, nitrogen in a diatomic molecular state as O₂, N₂ respectively, are the examples. .

20. How are compounds formed?

Answer: Molecules of compounds are created, when atoms of various elements combine with each other. This chemical attraction between different elements helps to form the compound.

21. What is Avogadro’s number?

Answer: It is seen that the number of molecules in one mole of any substance is constant. An Italian scientist, Avogadro did some pioneering work in this context, and thus this is known as Avogadro’s number denoted by the symbol NA, the value of Avogadro’s number is 6.022 x 10²³.

22. How many molecules are there in 66 g of CO₂?

Answer: Molecular mass of CO₂ is 44.

Hence, No of molecules(n) of CO₂ = mass of CO₂ in grams / molecular mass of CO₂

i.e., 66/44

Thus, n=1.5 mol. Therefore, if 1 mol consists of 6.022 x 10^23,molecules, then 1.5 mol contains = 6.022 x 10²³ x 1.5 = 9.033 x 10²³ molecules.

23. What are cations and anions? Give an example.

Answer: Cations are the positively charged ions while the negatively charged ions are termed anions. Here, in the example, MgCl₂ contains Mg⁺⁺ and Cl^– as cation and anion, respectively.

24. What are valence electrons?

Answer: The electrons found in the outermost orbit of an atom are called valence electrons.

25. What is the National Chemical Laboratory (NCL)?

Answer: The National Chemical Laboratory (NCL) is a unit of the CSIR and was set up in 1950. It aims to conduct research in the various branches of chemistry, aid industry and to develop new technology with a view to making profitable use of the country’s natural resources. The Laboratory conducts research in fields such as biotechnology, nanotechnology and polymer science.

26. What is variable valency?

Answer: The atoms of some elements give away or take up different numbers of electrons, in different conditions. This property of the elements of exhibiting more than one valency in such scenarios is called variable valency.