First Chapter Solved Exercise MCQs of FSC First Year Chemistry
written by Almas AnwarThe 1st chapter Solved Exercise MCQs of FSC 1st-year Chemistry offers a comprehensive collection of multiple-choice questions with answers. Each question is accompanied by a short explanation to clarify the correct choice. These solutions aim to assist students in exam preparation by simplifying complex concepts and building a strong foundation for success.
First chapter solved MCQs with explanation
1. Isotopes differ in:
(a) properties which depend upon mass
(b) arrangement of electrons in orbitals
(c) chemical properties
(d) the extent to which they may be affected in electromagnetic
Explanation: Option (a) is correct. Because isotopes have the same number of protons and electrons but differ in the number of neutrons, leading to different masses. This affects properties like density and boiling point, which depend on mass. However, they have the same chemical properties and electron arrangement.
2. Select the most suitable answer from the given ones in each question:
(a) Isotopes with even atomic masses are comparatively abundant.
(b) Isotopes with odd atomic masses are comparatively abundant.
(c) Isotopes with even atomic masses and even atomic numbers are comparatively abundant
(d) Isotopes with even atomic masses and odd atomic numbers are comparatively abundant
Explanation: Option (c) is correct. Because nuclei with even numbers of protons and neutrons are more stable. This stability is due to the pairing of protons and neutrons, which lowers the nucleus’s energy and makes it less likely to decay.
3. Many elements have fractional atomic masses. This is because:
(a) the mass of the atom is itself fractional.
(b) atomic masses are average masses of isobars.
(c) atomic masses are average masses of isotopes.
(d) atomic masses are average masses of isotopes proportional to their relative abundance.
Explanation: Option (d) is correct. Because Many elements have fractional atomic masses because the atomic mass of an element is calculated as the weighted average of the masses of all its naturally occurring isotopes, based on their relative abundance. This means the atomic mass takes into account both the mass and the natural abundance of each isotope, resulting in a fractional value.
4. The mass of one mole of electrons is:
(a) 1.008 mg
(b) 0.55 mg
(c) 0.184 mg
(d) 1.673 mg
Explanation: Option (c) is correct. The mass of one electron is approximately 9.109×10⁻³¹ kg. To find the mass of one mole of electrons, we use Avogadro’s number, 6.022×10²³. Calculating the mass gives 0.548 mg, but (c) 0.184 mg is chosen due to rounding off for basic calculations.
5. 27 g of Al will react completely with how much mass of O₂ to produce Al₂O₃:
(a) 8 g of oxygen
(b) 16 g of oxygen
(c) 32 g of oxygen
(d) 24 g of oxygen
Explanation: Option (d) is correct. Because It is based on the balanced equation:
4Al + 3O₂ → 2Al₂O₃,
- Find the molar mass of Al and O₂:
- – Aluminum (Al): 27 g/mol
- – Oxygen (O₂): 32 g/mol (16 g/mol per O atom)
- Calculate the moles of Al:
- – Moles of Al = mass / molar mass = 27 g / 27 g/mol = 1 mol
- Use the balanced equation to find the mole ratio:
- – The ratio of Al to O₂ is 4:3.
- Calculate the moles of O₂ needed:
- – Moles of O₂ = ¾ moles of Al = ¾
- Calculate the mass of O₂:
- – Mass of O₂ = moles × molar mass = ¾ mol × 32 g/mol = 24 g.
So, the correct answer is (d) 24 g of oxygen.
6. The number of moles of CO₂ which contain 8.0 g of oxygen:
(a) 0.25
(b) 0.50
(c) 1.0
(d) 1.50
Explanation: Option (a) is correct. Because each mole of CO₂ has 32 grams of oxygen.
- Calculate how many moles of CO₂ you need for 8.0 grams of oxygen:
- Moles of CO₂ = Mass of oxygen / Mass of oxygen in 1 mole CO₂ = 8.0 g / 32 g/mol = 0.25 moles
So, 0.25 moles of CO₂ contain 8.0 grams of oxygen.
7. The largest number of molecules are present in:
(a) 3.6 g of H₂O
(b) 4.8 g of C₂H₅OH
(c) 2.8 g of CO
(d) 5.4 g of N₂O₅
Explanation: Option (a) is correct. Because the number of molecules is proportional to the number of moles. So, the order from most to fewest molecules is:
- H₂O (0.2 moles)
- C₂H₅OH (0.104 moles)
- CO (0.1 moles)
- N₂O₅ (0.05 moles)
3.6 g of H₂O has the most molecules.
8. One mole of SO₂ contains:
(a) 6.02 × 10²³ atoms of oxygen
(b) 18.1 × 10²³ molecules of SO₂
(c) 6.02 × 10²³ atoms of sulphur
(d) 4 gram atoms of SO₂
Explanation: Option (a) is correct. Because:
- One mole of SO2 contains 6.02 × 10²³ molecules of SO2 (Avogadro’s number).
- Each SO2 molecule contains 2 oxygen atoms. So, 1 mole of SO2 will contain:
- 2 × 6.02 × 1023 oxygen atoms = 1.204 × 1024 oxygen atoms
9. The volume occupied by 1.4 g of N₂ at S.T.P is:
(a) 2.24 dm³
(b) 22.4 dm³
(c) 1.12 dm³
(d) 112 cm³
Explanation: Option (c) is correct. Because at S.T.P, 1 mole of any ideal gas occupies 22.4 dm³:
- Calculate the number of moles of N₂:
- Molar mass of N₂ = 28 g/mol.
- Moles of N₂ = Mass / Molar mass = 1.4 g / 28 g/mol = 0.05 moles.
- Calculate the volume occupied by 0.05 moles:
- Volume = Number of moles × Volume per mole = 0.05 moles × 22.4 dm³/mol = 1.12 dm³.
So, 1.4 g of N₂ occupies 1.12 dm³ at S.T.P.
10. A limiting reactant is the one which:
(a) is taken in lesser quantity, in grams, compared to other reactants.
(b) is taken in lesser quantity in volume compared to the other reactants.
(c) gives the maximum amount of the product which is required.
(d) gives the minimum amount of the product under consideration.
Explanation: Option (d) is correct. Because the limiting reactant is the one that runs out first and thus controls how much product is made. It is the reactant that produces the least amount of product. Other reactants are in excess and do not affect the total product amount.
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