Solved Examples: Some Basic Concepts of Chemistry | Physical Chemistry for NEET PDF Download

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Basic Concepts of Chemistry

1. The value of Avogadro constant is
(a) 6.022 x 10²² atoms
(b) 6.022 x 10²³ atoms
(c) 6.022 x 10²³ mol^–1
(d) 6.022 x 10²² mol^–1
Ans: c
Solution: Avogadro constant is 6.022 x 10²³ mol^–1.

2. The number of significant digits in the number 126000 is
(a) 3
(b) 4
(c) 5
(d) 6
Ans: a
Solution: There are 3 digits. The last three zeros are not significant.

3. The relative atomic mass of sodium is 23. Which of the following statements is not correct about sodium?
(a) Atomic mass of sodium is 23 u.
(b) Atomic mass of sodium is 3.82 x 10^-26 kg.
(c) Molar mass of sodium is 23 g mol^-1.
(d) The number of atoms in 24 kg of sodium is 6.022 x 10²³.
Ans: d
Solution: The number of atoms in 23 g of Na will be equal to 6.022 x 10²³

4. The radius of a hydrogen atom is 5.29 X 10^-11 m and that of a proton is 1.5 x10^-15 m. The ratio of the radius of atom to the radius of proton in the scientific notation will be
(a) 3.526 x 10⁴
(b) 35.266 x 10³
(c) 3.5 x 10⁴
(d) 3.526 6667 x10³
Ans: c
Solution:

5. Mass percent of Na in Na₂CO₃ is
(a) 21.52%
(b) 31.20%
(c) 38.20%
(d) 43.40%
Ans: d
Solution: Mass percent of Na

6. Select the quantity of NO₂ with highest mass:
(a) 100 amu
(b) 1.0 x 10^–3 g
(c) 7.0 x 10²² molecules
(d) 8.0 x 10^–1 mol
Ans: d
Solution: 100 amu = (100)

= 1.66 x 10^–22 g
Mass of 7.0 x 10²² molecules =
Mass of 8.0 x 10^–1 mol = 0.8 x 46 g = 36.8 g

7. The volume of 0.25 M NaOH to be added to 250 mL of 0.15 M NaOH so that the resultant solution is 0.2 M would be
(a) 250 mL
(b) 350 mL
(c) 450 mL
(d) 550 mL
Ans: a
Solution: Let V be the volume of 0.25 M NaOH solution. Total amount of NaOH after mixing the two solutions is n = V (0.25 mol L^–1) + (0.25 L) (0.15 mol L^–1)
Total volume of the solution = V + 0.25 L
Molarity of the resultant solution

Equating this to 0.2 M, we get

Solving for V, we get V = 0.25 L = 250 mL

8. The mass of H₂O₂ that is completely oxidised by 30.2 g of KMnO₄ (molar mass = 158 g mol^–1) in acidic medium is
(a) 12 g
(b) 15 g
(c) 17 g
(d) 1 g
Ans: c
Solution: The reaction is 2KMnO₄ + 5H₂O₂ + 6H⁺ → 2Mn²⁺ + 5O₂ + 8H₂O + 2K⁺
2 x 151 g of KMnO₄ reacts completely with 5 x 34 g of H₂O₂
30.2 g of KMnO₄ reacts completely with