NCERT MOST IMPORTANT QUESTIONS CLASS – 11 | CHEMISTRY IMPORTANT QUESTIONS PART-1 | CHAPTER -1 | SOME BASIC CONCEPTS OF CHEMISTRY | EDUGROWN |

In This Post we are  providing Chapter-1 SOME BASIC CONCEPTS OF CHEMISTRY NCERT MOST IMPORTANT QUESTIONS for Class 11 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON SOME BASIC CONCEPTS OF CHEMISTRY

Question 1.
Define Mole. What is its numerical value?

Answer:
A mole is the amount of a substance that contains as many entities (atoms, molecules, or other particles) as there are atoms in exactly 0.012 kg or 12 g of the carbon-12 isotope.

Its numerical value is 6.023 × 10²³.

Question 2.
Define molarity. Is it affected by a change in temperature?

Answer:
The molarity of a solution is defined as the number of moles of the solute present per liter of the solution. It is represented by the symbol M. Its value changes with the change in temperature.

Question 3.
What do you mean by Precision and accuracy?

Answer:
Precision and accuracy: The term precision refers to the closeness of the set of values obtained from identical measurements of a quantity.

Accuracy refers to the closeness of a single measurement to its true value.

Question 4.
Distinguish between fundamental and the derived units.

Answer:
Fundamental units: Fundamental units are those units by which other physical units can be derived. These are mass (M), Length (L), time (T), temperature (°).

Derived units: The units which are obtained by the combination of the fundamental units are called derived units.

Question 5.
Define molality and write its temperature dependence.

Answer:
Molality is defined as the number of g moles of the solute dissolved per kilogram of the solvent.
Molality (m) =  Mole of solute  Mass of the solvent in kg

The molality of the solution does not depend upon the temperature.

Question 6.
Two containers of equal capacity A₁ and A₂ contain 10 g of oxygen (O₂) and ozone (O₃) respectively. Which of the two will have greater no. of O-atoms and which will give greater no. of molecules?

Answer:
10 g of O₂ = 1032 mol = 1032 × 6.02 × 10²³ molecules
= 1.88 × 10²³ molecules
= 3.76 × 10²³ atoms.

10 g of O₂ = 1048 mole = 1048 × 6.02 × 10²³ molecules
= 1.254 × 10²³ molecules
= 3.76 × 10²³ atoms

Thus both A₁ and A₂ contain the same no. of atoms, but A₁ contains more numbers of molecules.

Question 7.
Assuming the density of water to be 1 g/cm³, calculate the volume occupied by one molecule of water.

Answer:
1 Mole of H₂O = 18 g = 18 cm³[∵ density of H₂O = 1 g/cm³]
= 6.022 × 10²³ molecules of H₂O
1 Molecule will have a volume
= 186.022×1023 cm- = 2.989 × 10^-23 cm³.

Question 8.
State the law of Multiple Proportions. Explain with two examples.

Answer:
The Law of Multiple Proportions states:
“When two elements combine to form two or more than two chemical compounds than the weights of one of elements which combine with a fixed weight of the other, bear a simple ratio to one another.

Examples:
1. Compound of Carbon and Oxygen: C and O combine to form two compounds CO and CO₂.
In CO₂ 12 parts of wt. of C combined with 16 parts by wt. O.
In CO₂ 12 parts of wt. of C combined with 32 parts by wt. of O.
If the weight of C is fixed at 12 parts by wt. then the ratio in the weights of oxygen which combine with the fixed wt. of C (= 12) is 16: 32 or 1: 2.
Thus the weight of oxygen bears a simple ratio of 1: 2 to each other.

2. Compounds of Sulphate (S) and Oxygen (O):
S forms two oxides with O, viz., SO₂ and SO₃
In SO₂, 32 parts of wt. of S combine with 32 parts by wt. of O.
In SO₃, 32 parts of wt. of S combine with 48 parts by wt. of O.
If the wt. of S is fixed at 32 parts, then’ the ratio in the weights of oxygen which combine with the fixed wt. of S is 32: 48 or 2: 3.

Thus the weights of oxygen bear a simple ratio of 2: 3 to each other.

Question 9.
State the law of Constant Composition. Illustrate with two examples.

Answer:
Law of Constant Composition of Definite Proportions states: “A chemical compound is always found, to be made up of the same elements combined together in the same fixed proportion by weight”.

Examples:

1. CO₂ may be prepared in the laboratory as follows:

In all the above examples, CO₂ is made up of the same elements i. e., Carbon (C) and Oxygen (O) combined together in the same fixed proportion by weight of 12: 32 or 3: 8 by weight.

Question 10.
Define empirical formula and molecular formula. How will you establish a relationship between the two? Give examples.

Answer:
The empirical formula of a compound expresses the simplest whole-number ratio of the atoms of the various elements present in one molecule of the compound.

For example, the empirical formula of benzene is CH and that of glucose is CH₂O. This suggests that in the molecule of benzene one atom of Carbon (C) is present for every atom of Hydrogen (H). Similarly in the molecule of glucose (CH₂O), for every one atom of C, there are two atoms of H and one atom of O present in its molecule. Thus, the empirical formula of a compound represents only the atomic ratio of various elements present in its molecule.

The molecular formula of a compound represents the true formula of its molecule. It expresses the actual number of atoms of various elements present in one molecule of a compound. For example, the molecular formula of benzene is C₆H₆ and that of glucose is C₆H₁₂O₆. This suggests that in one molecule of benzene, six atoms of C and 6 atoms of H are present. Similarly, one molecule of glucose (C₆H₁₂O₆) actually contains 6 atoms of C, 12 atoms of H, and 6 atoms of O.

Relation between the empirical and molecular formula
Molecular formula = n × Empirical formula where n is an integer such as 1, 2, 3…
When n = 1; Molecular formula = Empirical formula
When n = 2; Molecular formula = 2 × Empirical formula.
The value of n can be obtained from the relation.
n =  Molecular mass  Empirical formula mass 

The molecular mass of a volatile substance can be determined by Victor Meyer’s method or by employing the relation.
Molecular mass = 2 × vapour density .

Empirical formula mass can however be obtained from its empirical formula simply by adding the atomic masses of the various atoms present in it.

Thus the empirical formula mass of glucose CH20
= 1 × 12 + 2 × 1 + 1 × 16 = 30.0 u.