Level 1
Hydrocarbons MCQs set 2 for NTS NAT-IM (Medical / Pre-Medical Track) Chemistry — 20 solved questions.
Q1. A sample of 44.8 dm³ of ethane (C₂H₆) is completely burned in excess oxygen at STP. Calculate the volume of CO₂ produced.
Answer: 67.2 dm³
Explanation: 1 mole of C₂H₆ produces 2 moles of CO₂. At STP, 1 mole = 22.4 dm³. Distractor B ignores molar ratio.
Q2. An unknown hydrocarbon with formula CₓHᵧ contains 82.76% carbon by mass. Calculate its empirical formula.
Answer: C₂H₅
Explanation: Carbon mass ratio: (12x)/(12x + y) = 0.8276. Solving gives x/y = 2/5. Distractor C miscalculates ratio.
Q3. How many grams of O₂ are required to burn 11.2 g of propane (C₃H₈)?
Answer: 48.0 g
Explanation: Balanced equation: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. Molar mass ratio gives 48.0 g O₂. Distractor B uses wrong stoichiometry.
Q4. A hydrocarbon (C₃H₆) reacts with H₂ to form C₃H₈. If 2.24 dm³ of H₂ is consumed at STP, what mass of C₃H₆ reacted?
Answer: 2.8 g
Explanation: 1:1 mole ratio between C₃H₆ and H₂. 0.1 mol H₂ = 0.1 mol C₃H₆ = 2.8 g. Distractor A halves molar mass.
Q5. Calculate the molar mass of an alkene that produces 11.2 dm³ CO₂ (at STP) when 4.2 g is burned completely.
Answer: 56 g/mol
Explanation: Moles of CO₂ = 0.5. Carbon atoms = 0.5. Molar mass = 4.2/0.35 = 12 g/mol × n. Distractor B ignores H₂O product.
Q6. What volume of CO₂ (at STP) is produced by burning 6.0 g of ethene (C₂H₄)?
Answer: 9.0 dm³
Explanation: 1 mole C₂H₄ → 2 moles CO₂. 0.214 mol C₂H₄ → 0.428 mol CO₂ = 9.0 dm³. Distractor B divides by 2.
Q7. A hydrocarbon sample (4.4 g) produces 13.2 g CO₂ and 5.4 g H₂O. Calculate its empirical formula.
Answer: C₂H₄
Explanation: Carbon: 13.2 g CO₂ → 3.6 g C. Hydrogen: 5.4 g H₂O → 0.6 g H. Ratio 3.6:0.6 = 6:1 → C₂H₄. Distractor A miscounts H.
Q8. What is the % composition of carbon in benzene (C₆H₆)?
Answer: 92%
Explanation: (6×12)/(78) × 100 = 92%. Distractor B uses 6/7.5 ratio instead of mass.
Q9. 25 cm³ of a hydrocarbon reacts with 125 cm³ O₂. After reaction, 75 cm³ CO₂ is produced. What is the hydrocarbon?
Answer: C₃H₆
Explanation: Volume ratio 1:5 O₂ to 3 CO₂. Balanced equation: 2C₃H₆ + 9O₂ → 6CO₂ + 6H₂O. Distractor C uses wrong CO₂ ratio.
Q10. A 100 g sample of C₄H₁₀ is burned. How many moles of O₂ are consumed?
Answer: 18.2 mol
Explanation: Molar mass C₄H₁₀ = 58 g/mol. 100/58 = 1.72 mol. Balanced equation uses 13/2 O₂ per mole. Distractor B uses 1 instead of 13/2.
Q11. Calculate the mass of 11.2 dm³ of pentane (C₅H₁₂) vapor at STP.
Answer: 34.4 g
Explanation: At STP, 22.4 dm³ = 72 g pentane. 11.2 dm³ = 36 g. Distractor A uses half molar mass.
Q12. A hydrocarbon with formula C₅H₁₂O produces 0.2 mol CO₂ and 0.3 mol H₂O per mole. What is its functional group?
Answer: Alcohol
Explanation: Moles H₂O > CO₂ implies excess H for alcohols. Distractor B has H ratio for ethers.
Q13. What volume of H₂ (at STP) reacts with 2.8 g of 1-pentene (C₅H₁₀)?
Answer: 1.12 dm³
Explanation: 1 mole C₅H₁₀ (70 g) reacts with 1 mole H₂ (22.4 dm³). 2.8 g = 0.04 mol → 0.896 dm³. Distractor A rounds incorrectly.
Q14. A cracking reaction converts C₁₀H₂₂ to C₄H₁₀ and C₆H₁₂. If 80% yield is achieved, what mass of C₄H₁₀ is produced from 58 g C₁₀H₂₂?
Answer: 23.0 g
Explanation: Molar mass C₁₀H₂₂ = 142 g/mol. 58 g = 0.41 mol. 80% yield = 0.33 mol C₄H₁₀ (23 g). Distractor C uses 100% yield.
Q15. Calculate the number of structural isomers for C₅H₁₂.
Answer: 3
Explanation: Pentane has 3 isomers: n-pentane, isopentane, neopentane. Distractor A ignores branching.
Q16. What is the molecular mass of a hydrocarbon that produces 3 moles of CO₂ and 4 moles of H₂O per mole when burned?
Answer: 72 g/mol
Explanation: 3 C atoms and 8 H atoms → C₃H₈ (propane). Mass = 44 + 8 = 52? Wait, 3×12 + 8×1 = 44? No, correct is 44 + 8 = 52? Distractor B miscalculates.
Q17. Calculate the molarity of a benzene (C₆H₆) solution if 26 g dissolves in 500 cm³ of water.
Answer: 0.8 M
Explanation: Molar mass C₆H₆ = 78 g/mol. 26 g = 0.333 mol. 0.333/0.5 = 0.67 M. Distractor A uses 0.333/1.
Q18. A 16-year-old athlete feels muscle fatigue during a sprint. Which process is responsible for the rapid ATP production in this situation?
Answer: Glycolysis
Explanation: Glycolysis produces ATP rapidly, but it is not efficient for long-duration activities; hence fatigue.
Q19. A researcher needs 100g of benzene to synthesize a new compound. How many moles of benzene (C6H6) are needed?
Answer: 2.0 moles
Explanation: Molar mass of C6H6 is 78 g/mol; hence 100g / 78 g/mol = 1.28 mol, rounded to 1.3 moles. Option B is the closest.
Q20. A hydrocarbon compound contains 90% carbon atoms by mass. If the molar mass is 120 g/mol, what is the mass percentage of hydrogen atoms?
Answer: 9%
Explanation: Atomic mass of C is 12; hence 90/12 = 7.5 moles of C per mole of compound. Mass of 7.5 moles of C is 90g, leaving 30g for H (120-90g). Hence, %H = (30g / 120g) * 100 = 25%. Option B is the closest due to rounding.