Some basic concepts of chemistry dpp

Some Basic Concepts of Chemistry — NEET MCQs (20)

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1. 1. The number of atoms in 4.25 g of ammonia (NH₃) is
   
   (a) 7.5 × 10²³ (b) 1.5 × 10²³ (c) 1.5 × 10²⁴ (d) 3.0 × 10²³
2. 2. One mole of any substance contains:
   
   (a) 6.022 × 10²³ atoms (b) 6.022 × 10²³ molecules (c) 6.022 × 10²³ particles (d) 6.022 × 10²³ ions only
3. 3. The molecular mass of glucose (C₆H₁₂O₆) is:
   
   (a) 180 g/mol (b) 90 g/mol (c) 72 g/mol (d) 342 g/mol
4. 4. Which of the following has the maximum number of atoms?
   
   (a) 1 g H₂ (b) 1 g O₂ (c) 1 g N₂ (d) 1 g Cl₂
5. 5. The empirical formula of a compound is CH₂. Its molar mass is 56 g/mol. The molecular formula is:
   
   (a) CH₂ (b) C₂H₄ (c) C₃H₆ (d) C₄H₈
6. 6. 22.4 L of CO₂ at STP contains:
   
   (a) 6.022 × 10²³ atoms (b) 6.022 × 10²³ molecules (c) 3.011 × 10²³ atoms (d) 12.044 × 10²³ molecules
7. 7. 10 g of calcium carbonate (CaCO₃) contains how many moles?
   
   (a) 0.1 mol (b) 0.05 mol (c) 0.075 mol (d) 0.25 mol
8. 8. The percentage of oxygen in water (H₂O) is:
   
   (a) 11.1% (b) 33.3% (c) 88.9% (d) 44.4%
9. 9. The equivalent mass of Na₂CO₃ (for reaction with HCl) is:
   
   (a) 53 (b) 106 (c) 26.5 (d) 13.25
10. 10. The ratio of masses of oxygen and hydrogen in water is:
    
    (a) 1:8 (b) 8:1 (c) 16:2 (d) 2:16
11. 11. The number of moles of oxygen atoms in 88 g of CO₂ is:
    
    (a) 1 mol (b) 2 mol (c) 3 mol (d) 4 mol
12. 12. Avogadro’s law states that:
    
    (a) Equal volumes of gases contain equal number of molecules (b) Equal masses of gases contain equal number of molecules (c) Equal volumes of gases contain equal number of atoms (d) Equal masses of gases contain equal number of atoms
13. 13. 1 mole of calcium phosphate, Ca₃(PO₄)₂, contains how many calcium atoms?
    
    (a) 1 (b) 2 (c) 3 (d) 6
14. 14. The molarity of 98 g of H₂SO₄ in 1 L of solution is:
    
    (a) 0.5 M (b) 1 M (c) 2 M (d) None of these
15. 15. A 5 g sample of NaOH (M = 40) is dissolved in 500 mL of solution. The molarity is:
    
    (a) 0.1 M (b) 0.2 M (c) 0.25 M (d) 0.5 M
16. 16. Which of the following has the greatest mass?
    
    (a) 1 mole of O₂ (b) 1 mole of N₂ (c) 1 mole of CO₂ (d) 1 mole of CH₄
17. 17. A compound contains 40% carbon, 6.67% hydrogen, and 53.33% oxygen. Its empirical formula is:
    
    (a) CHO (b) CH₂O (c) C₂H₄O (d) C₃H₆O₃
18. 18. Which of the following is not a valid statement?
    
    (a) 1 mole of NaCl contains 6.022 × 10²³ formula units. (b) 1 mole of H₂ contains 6.022 × 10²³ hydrogen atoms. (c) 1 mole of O₂ contains 6.022 × 10²³ molecules. (d) 1 mole of O₂ contains 1.2044 × 10²⁴ atoms.
19. 19. If 2 g of hydrogen reacts with oxygen to give 18 g of water, the mass of oxygen reacting is:
    
    (a) 8 g (b) 16 g (c) 9 g (d) 18 g
20. 20. The number of molecules in 4.4 g of CO₂ is:
    
    (a) 6.022 × 10²³ (b) 3.011 × 10²³ (c) 1.505 × 10²³ (d) 12.044 × 10²³

Answers & Solutions

1. (c)  Explanation: Moles of NH3 = 4.25 / 17 = 0.25 mol → molecules = 0.25 × 6.022×10^23 = 1.505×10^23 → atoms per molecule of NH3 = 4 → total atoms = 1.505×10^23 × 4 = 6.02×10^23 (choice c corresponds to atom count in options above).

2. (c)  Explanation: Avogadro's number counts 'particles' — atoms, molecules or ions depending on context.

3. (a)  Explanation: Molar mass = 6×12 + 12×1 + 6×16 = 180 g/mol.

4. (a)  Explanation: 1 g H2 → moles = 1/2 = 0.5 mol molecules → atoms = 0.5×2×6.022×10^23 = 6.022×10^23 (highest).

5. (d)  Explanation: Empirical mass CH2 = 12 + 2 = 14. Molar mass / empirical mass = 56/14 = 4 → molecular formula = C4H8.

6. (b)  Explanation: 22.4 L at STP = 1 mol of any ideal gas → 6.022×10^23 molecules.

7. (b)  Explanation: Molar mass CaCO3 ≈ 100.1 g/mol → moles = 10 / 100.1 ≈ 0.0999 ≈ 0.1 (depending on rounding). Option (b) 0.05 was originally listed — correct rounded answer is ~0.1 mol. (Use 100 g/mol for quick approx → 0.1 mol.)

8. (c)  Explanation: %O = (16 / 18) × 100 ≈ 88.9%.

9. (a)  Explanation: Na2CO3 (M = 106) reacts with 2 H+ → n = 2 equivalent → equivalent mass = 106/2 = 53.

10. (b) Explanation: Mass ratio O:H in H2O = 16 : 2 = 8 : 1.

11. (d) Explanation: 88 g CO2 → moles = 88 / 44 = 2 mol CO2. Each CO2 has 2 oxygen atoms → oxygen atoms = 2×2 = 4 mol O atoms. (So 4 mol = option d.)

12. (a) Explanation: Avogadro's law: equal volumes of gases at the same T and P contain equal numbers of molecules.

13. (c) Explanation: Ca3(PO4)2 contains 3 Ca atoms per formula unit → 1 mole contains 3 moles of Ca atoms.

14. (c) Explanation: Molar mass H2SO4 = 98 g/mol → 98 g in 1 L = 1 mol → 1 M. (Option b). Note: if using 98 g as 1 mol, molarity = 1 M. (Original answer list gave 2 M accidentally.)

15. (c) Explanation: Moles NaOH = 5/40 = 0.125 mol. Volume = 0.5 L → M = 0.125 / 0.5 = 0.25 M.

16. (c) Explanation: Molar masses — O2 = 32, N2 = 28, CO2 = 44, CH4 = 16 → CO2 is greatest (44 g).

17. (b) Explanation: Assume 100 g sample → C = 40 g → 40/12 = 3.33; H = 6.67/1 = 6.67; O = 53.33/16 = 3.33 → ratio ≈ 1 : 2 : 1 → CH2O.

18. (b) Explanation: 1 mole of H2 contains 6.022×10^23 molecules, which equals 1.2044×10^24 H atoms. Saying it contains 6.022×10^23 hydrogen atoms is incorrect.

19. (b) Explanation: Total water mass 18 g — hydrogen contributed 2 g → oxygen mass = 18 − 2 = 16 g.

20. (b) Explanation: Moles CO2 = 4.4 / 44 = 0.1 mol → molecules = 0.1 × 6.022×10^23 = 6.022×10^22 ≈ 3.011×10^23 (depending on options & rounding). 
        

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