Some Basic Concepts of Chemistry Class 11 – MCQs (Answers)

Ans 1.

Ans 2. 6.023 × 10²³ has four significant figures. The total number of digits in a number including the last digit whose value is uncertain is called the number of significant figures.

Ans 3. All non-zero digits are significant. Hence, there are 6 significant figures in 10.3406 g.

Ans 4. 0.02856 × 298.15 × 0.112 /0.5785

Ans 5. Precision refers to the closeness of various measurements for the same quantity. Accuracy is the agreement of a particular value to the true value of the result. Thus, A and B are precision and accuracy respectively.

Ans 6. As we know, the least count of an instrument is equal to the most possible error of the instrument, hence most possible error of the instrument will be 0.01 cm.

Ans 7. All the given statements are correct about molecular mass.

Ans 8. Molecular mass of urea (NH₂CONH₂) is
= 2 × (Atomic mass of N) + 4 × (Atomic mass of H) + (Atomic mass of C) + (Atomic mass of O) = 60
Mass percentage of N = (28/60) × 100 = 46.67 %

Ans 9.

Ans 10. One mole of any substance contains 6.022 × 10²³ atoms/ molecules. Hence, the number of millimoles of H₂SO₄ = Molarity × Volume (in mL) = 0.02 × 100 = 2 × 10⁻³ mol
Number of molecules = number of moles × N_A
= 2 × 10⁻³ × 6.022 × 10²³ = 12.044 × 10²⁰ molecules

Ans 11. In 18 g of water, the number of H₂O molecules is N_A .
18 mg of water, the number of H₂O molecules would be =

                                                                                            = 10^-3 × N_A

Ans 12. As 22400 mL water contains water molecules = 6.023 × 10²³ molecules

Ans 13.

Ans 14.

Ans 15. Molecular mass of CO₂ = 1 × 12 + 2 × 16 = 44 g
1 g molecule of CO₂ contains 1 g atoms of carbon
؞ 44 g of CO₂ contain C = 12 g atoms of carbon
% of C in CO₂ = 12/44 × 100 = 27.27 %

Ans 16.
Ans 17.

Ans 18. 2Na₂HPO₄ + NaH₂PO₄ + 2(NH₂)₂ CO → Na₅P₃O₁₀ + 4NH₃ + 2CO₂
Hence, the stoichiometric ratio of sodium dihydrogen orthophosphate and sodium hydrogen orthophosphate is 2:1

Ans 19. The molar mass of sulphuryl chloride is 32 + 16 + 16 + 35.5 + 35.5 = 135 g/mol
The number of gram molecules in 13.5 g of sulphuryl chloride is = 13.5/135 = 0.1 mol

Ans 20. One litre of the solution

Ans 21. (e) the assertion is false but the reason is true

Ans 22. (e) the assertion is false but the reason is true

Ans 23. (c) the assertion is true but the reason is false.

Ans 24. (e) the assertion is false but the reason is true

Ans 25. (e) the assertion is false but the reason is true