Third Chapter Solved Exercise MCQs of FSC First Year Chemistry
written by Almas AnwarThe 3rd 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.
Third chapter solved MCQs with explanation
1. Pressure remaining constant, at which temperature will the volume of a gas become twice what it is at 0°C?
(a) 546°C
(b) 200°C
(c) 546K
(d) 273K
Explanation: The correct answer is (c). Because Charles’s Law states that the volume of a gas is directly proportional to its temperature when the pressure is constant. To find the temperature at which the volume of the gas becomes twice its original volume at 0°C, we use the relationship:
- At 0°C, the temperature in Kelvin is 273K.
- To double the volume, the temperature also needs to double in Kelvin. So, the new temperature should be: 273K×2 = 546K. Thus, the volume of the gas will become twice at 546K.
2. The number of molecules in one dm³ of water is close to:
(a) 6.02/22.4 x 10²³
(b) 12.04/22.4 x 10²³
(c) 18/22.4 x 10²³
(d) 55.6 x 6.02 x 10²³
Explanation: The correct answer is (d). Because:
- Density of water: 1 dm³ of water weighs 1000 grams.
- Molar mass of water: 18 grams per mole.
- Number of moles in 1000 grams: 1000/18 ≈ 55.6 moles 181000 ≈ 55.6 moles
- Molecules in 55.6 moles: 55.6 × 6.02 × 1023
3. Which of the following will have the same number of molecules at STP?
(a) 280 cm³ of CO₂ and 280 cm³ of N₂O
(b) 11.2 dm³ of O and 32 g of O₂
(c) 44 g of CO₂ and 11.2 dm³ of CO
(d) 28 g of N₂ and 5.6 dm³ of oxygen
Explanation: The correct answer is (a). Because:
- At STP : 1 mole of any gas occupies 22,400 cm³ and contains the same number of molecules.
- Option (a): Both CO₂ and N₂O have the same volume (280 cm³) at STP, so they have the same number of molecules.
- Other options either have different volumes or different masses. so they do not have the same of molecules.
4. If the absolute temperature of a gas is doubled and the pressure is reduced to one half, the volume of the gas will:
(a) remain unchanged
(b) increase four times
(c) reduce to ¼
(d) be doubled
Explanation: The correct answer is (b). Because this problem involves the combined gas law, which relates pressure (P), volume (V), and temperature (T) of a gas:
P₁V₁ / T₁ = P₂V₂ / T₂.
- Given:
- – Temperature is doubled: T₂ = 2T₁
- – Pressure is halved: P₂ = P₁ / 2.
- Substitute these into the equation:
- – P₁V₁ / T₁ = (P₁ / 2)V₂ / 2T₁.
- Simplify:
- –V₂ = 4V₁, so the volume V₂ increases four times compared to the original volume V₁.
5. How should the conditions be changed to prevent the volume of a given gas from expanding when its mass is increased?
(a) Temperature is lowered and pressure is increased.
(b) Temperature is increased and pressure is lowered.
(c) Temperature and pressure both are lowered.
(d) Temperature and pressure both are increased.
Explanation: The correct answer is (a). Because lowering the temperature slows down gas particles, and increasing the pressure pushes them closer together, both of which prevent the gas from expanding.
6. The molar volume of CO₂ is maximum at:
(a) STP
(b) 127°C and 1 atm
(c) 0°C and 2 atm
(d) 273°C and 2 atm
Explanation: The correct answer is (b). Because:
- According to the ideal gas equation:
pV = nRT. (For n = 1, V = RT / p)
- For maximum volume, we need maximum volume of (T/p).
At NTP: T/p = 273K/1atm = 273Katm⁻¹.
At 0°C and 2atm: T/p = 273K/2atm = 136.5Katm⁻¹.
At 127°C and 1atm: T/p = 400K/1atm = 400Katm⁻¹.
At 273°C and 2atm: T/p = 546K/2atm = 273Katm⁻¹.
Thus, the molar volume of CO₂ is maximum at 127°C and 1atm.
7. The order of the rate of diffusion of gases NH₃, SO₂, Cl₂, and CO₂ is:
(a) NH₃ > SO₂ > Cl₂ > CO₂
(b) NH₃ > CO₂ > SO₂ > Cl₂
(c) Cl₂ > SO₂ > CO₂ > NH₃
(d) NH₃ > CO₂ > Cl₂ > SO₂
Explanation: The correct answer is (b). Because gases with lighter molecules diffuse faster than gases with heavier molecules. Molar masses of the gases: NH₃ (17 g/mol), CO₂ (44 g/mol), SO₂ (64 g/mol), Cl₂ (71 g/mol). Since NH₃ is the lightest, it diffuses the fastest. The correct order from fastest to slowest diffusion is NH₃ > CO₂ > SO₂ > Cl₂.
8. Equal masses of methane and oxygen are mixed in an empty container at 25°C. The fraction of total pressure exerted by oxygen is:
(a) 1/3
(b) 8/9
(c) 1/9
(d) 16/17
Explanation: The correct answer is (a). Methane (CH₄) and oxygen (O₂) are mixed in equal masses.
- Moles Calculation:
– Methane: Molar mass = 16 g/mol
– Oxygen: Molar mass = 32 g/mol
More moles of methane than oxygen (since methane has a lower molar mass).
- Partial Pressure:
-More moles mean more partial pressure.
-Calculate the fraction of the total pressure from oxygen using the ratio of moles.
Result: The fraction of the total pressure from oxygen is 1/3.
9. Gases deviate from ideal behavior at high pressure. Which of the following is correct for non-ideality?
(a) At high pressure, the gas molecules move in one direction only.
(b) At high pressure, the collisions between the gas molecules are increased manifold.
(c) At high pressure, the volume of the gas becomes insignificant.
(d) At high pressure, the intermolecular attractions become significant.
Explanation: The correct answer is (d). Because at high pressure, gas molecules are compressed into a smaller volume, so they are closer together. This close proximity increases the effect of intermolecular attractions between the molecules, which causes gases to deviate from ideal behavior.
10. The deviation of a gas from ideal behavior is maximum at:
(a) -10°C and 5.0 atm
(b) -10°C and 2.0 atm
(c) 100°C and 2.0 atm
(d) 0°C and 2.0 atm
Explanation: The correct answer is (a). Because the deviation of a gas from ideal behavior is maximum at high pressures and low temperatures. At high pressures, gas particles are closer together, increasing the effects of intermolecular forces. At low temperatures, gas particles move more slowly, making intermolecular forces more significant.
11. A real gas obeying van der Waals equation will resemble an ideal gas if:
(a) both ‘a’ and ‘b’ are large
(b) both ‘a’ and ‘b’ are small
(c) ‘a’ is small and ‘b’ is large
(d) ‘a’ is large and ‘b’ is small
Explanation: The correct answer is (b). Because in the van der Waals equation, (P + a/V²)(V – b) = RT, ‘a’ represents the strength of attractive forces between gas molecules, and ‘b’ represents the volume occupied by the gas molecules themselves. For a gas to behave like an ideal gas:
- ‘a’ should be small (no intermolecular attractions)
- ‘b’ should be small (volume of gas molecules is negligible)
So, when both ‘a’ and ‘b’ are small, the gas’s behavior will closely resemble that of an ideal gas.
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