Mole & Gases MCQ's PDF

1. What is the volume of 1 mole of any gas at room temperature and pressure (RTP)? a) 22.4 dm³ b) 24.0 dm³ c) 12.0 dm³ d) 21.2 dm³ Answer: b) 24.0 dm³ Explanation: At RTP (298 K and 1 atm), the molar gas volume is 24.0 dm³. 2. Calculate the number of moles in 48 dm³ of oxygen gas a...

1. What is the volume of 1 mole of any gas at room temperature and pressure (RTP)? a) 22.4 dm³ b) 24.0 dm³ c) 12.0 dm³ d) 21.2 dm³ Answer: b) 24.0 dm³ Explanation: At RTP (298 K and 1 atm), the molar gas volume is 24.0 dm³. 2. Calculate the number of moles in 48 dm³ of oxygen gas at RTP. a) 1 mole b) 2 moles c) 3 moles d) 4 moles Answer: b) 2 moles Explanation: Number of moles \( = \frac{\text{volume}}{\text{molar gas volume}} = \frac{48 \text{ dm}³}{24 \text{ dm}³/\text{mol}} = 2 \text{ mol} \). 3. What is the volume of 0.5 moles of CO\(_2\) at RTP? a) 12 dm³ b) 24 dm³ c) 48 dm³ d) 30 dm³ Answer: a) 12 dm³ Explanation: Volume \( = \text{number of moles} \times \text{molar gas volume} = 0.5 \times 24 \text{ dm}³/\text{mol} = 12 \text{ dm}³ \). 4. What volume of gas in cm³ is occupied by 0.1 moles of nitrogen at RTP? a) 2.40 × 10² cm³ b) 2.40 × 10³ cm³ c) 2.40 × 10⁴ cm³ d) 2.40 × 10⁵ cm³ Answer: c) 2.40 × 10⁴ cm³ Explanation: Volume \( = \text{number of moles} \times \text{molar gas volume} = 0.1 \times 24 \text{ dm}³ \times 1000 = 2.40 \times 10⁴ \text{ cm}³ \). 5. If 1.5 moles of a gas occupies 36 dm³, what is the molar volume of the gas? a) 18 dm³/mol b) 24 dm³/mol c) 36 dm³/mol d) 12 dm³/mol Answer: b) 24 dm³/mol Explanation: Molar volume \( = \frac{\text{volume}}{\text{number of moles}} = \frac{36 \text{ dm}³}{1.5 \text{ mol}} = 24 \text{ dm}³/\text{mol} \). 6. Calculate the volume of 2.5 moles of hydrogen gas at RTP. a) 60 dm³ b) 48 dm³ c) 24 dm³ d) 6 dm³ Answer: a) 60 dm³ Explanation: Volume \( = \text{number of moles} \times \text{molar gas volume} = 2.5 \times 24 \text{ dm}³/\text{mol} = 60 \text{ dm}³ \). 7. What volume is occupied by 4.8 g of oxygen gas (molar mass = 32 g/mol) at RTP? a) 56 dm³ b) 6 dm³ c) 3.6 dm³ d) 4.8 dm³ Answer: c) 3.6 dm³ Explanation: Number of moles \( = \frac{\text{mass}}{\text{molar mass}} = \frac{4.8 \text{ g}}{32 \text{ g/mol}} = 0.15 \text{ mol} \); Volume \( = 0.15 \times 24 \text{ dm}³ = 3.6 \text{ dm}³ \). 8. What is the pressure exerted by 2 moles of gas occupying 12 m³ at 300 K? (Use R = 8.31 J/K/ mol) a) 500 Pa b) 415 Pa c) 4150 Pa d) 8300 Pa Answer: d) 415 Pa Explanation: Using \( pV = nRT \); p \( = \frac{nRT}{V} = \frac{(2)(8.31)(300)}{12} = 415 \text{ Pa} \). 9. What is the volume of gas at 400 K and 120 kPa if there are 0.5 moles? (Use R = 8.31 J/K/mol) a) 16.625 m³ b) 13.85 m³ c) 0.138 m³ d) 1.385 m³ Answer: c) 0.138 m³ Explanation: Using \( pV = nRT \); V \( = \frac{nRT}{p} = \frac{(0.5)(8.31)(400)}{120 \times 1000} = 0.138 \text{ m}³ \). 10. Convert 20°C into Kelvin. a) 253 K b) 283 K c) 293 K d) 313 K Answer: c) 293 K Explanation: Temperature in Kelvin \( = \text{Temp in °C} + 273 = 20 + 273 = 293 \) K. 11. If the pressure is 300 kPa, convert this to Pascals (Pa). a) 3.00 × 10² Pa b) 3.00 × 10³ Pa c) 3.00 × 10⁴ Pa d) 3.00 × 10⁵ Pa Answer: d) 3.00 × 10⁵ Pa Explanation: Pressure in Pascals = Pressure in kPa × 1000 = 300 × 1000 = 3.00 × 10⁵ Pa. 12. Calculate the number of moles in 400 cm³ of nitrogen gas at RTP. a) 0.0167 moles b) 1.67 moles c) 0.167 moles d) 0.0375 moles Answer: d) 0.0167 moles Explanation: V \( = 400 \text{ cm}³ \times \frac{1 \text{ dm}³}{1000 \text{ cm}³} = 0.4 \text{ dm}³ \); n = \(\frac{V}{24} = \frac{0.4}{24} \approx 0.0167\) moles. 13. What is the volume in dm³ of 0.75 moles of methane gas at RTP?  a) 12 dm³ b) 24 dm³ c) 18 dm³ d) 6 dm³ Answer: c) 18 dm³ Explanation: Volume = \( n \times \text{molar gas volume} = 0.75 \times 24 = 18 \text{ dm}³ \). 14. Which is the correct formula for calculating volume using the ideal gas equation? a) V = \[ \frac{nRT}{p} \] b) V = \[ \frac{p}{nRT} \] c) V = \[ \frac{n}{RTp} \] d) V = \[ \frac{T}{npR} \] Answer: a) V = \[ \frac{nRT}{p} \] Explanation: Rearrange \[ pV = nRT \] to solve for volume, \[ V = \frac{nRT}{p} \)【0†source】. 15. Find the pressure in Pa if 3 moles of gas occupy 2.5 m³ at 350 K. (Use R = 8.31 J/K/mol) a) 3483.3 Pa b) 3943.6 Pa c) 3979.2 Pa d) 9362.2 Pa Answer: b) 3115.2 Pa Explanation: Use \( pV = nRT \); \( p = \frac{nRT}{V} = \frac{(3)(8.31)(350)}{2.5} = 3483.3 \text{ Pa} \). 16. What volume of H\(_2\) gas (in dm³) is produced from 1 mole of Mg reacting with HCl at RTP? a) 24 dm³ b) 12 dm³ c) 48 dm³ d) 6 dm³ Answer: b) 24 dm³ Explanation: 1 mole of Mg produces 1 mole of H\(_2\) gas. At RTP, 1 mole of any gas occupies 24 dm³. 17. How many moles of gas are present in a 30 dm³ container at 200 K and 2.5 atm? (Use R = 0.0821 atm·L/mol·K) a) 3.75 moles b) 4.57 moles c) 2.19 moles d) 6.15 moles Answer: a) 3.75 moles Explanation: Use \( pV = nRT \); \( n = \frac{pV}{RT} = \frac{(2.5)(30)}{0.0821 \times 200} = 3.75 \text{ moles} \). 18. Convert 5000 cm³ to m³. a) 5 × 10⁻³ m³ b) 5 × 10⁻² m³ c) 5 × 10⁻¹ m³ d) 5 × 10 m³ Answer: a) 5 × 10⁻³ m³ Explanation: Volume in m³ = Volume in cm³ × \(10^{-6}\); thus \( 5000 \text{ cm}³ = 5000 \times 10^{-6} = 5 \times 10^{-3} \text{ m}³ \). 19. What is the temperature in Kelvin if 3 moles of gas exert a pressure of 200 kPa in a 40 m³ container? (Use R = 8.31 J/K/mol)  a) 1287 K b) 2215 K c) 3623 K d) 2415 K Answer: a) 1287 K Explanation: Using \( pV = nRT \); \( T = \frac{pV}{nR} = \frac{(200 \times 10^3)(40)}{(3)(8.31)} = 1287 \text{ K} \). 20. Calculate the number of moles in 10 dm³ of chlorine gas at RTP. a) 0.25 moles b) 0.42 moles c) 0.83 moles d) 1.01 moles Answer: c) 0.42 moles Explanation: Number of moles = \(\frac{\text{Volume}}{\text{Molar gas volume}} = \frac{10 \text{ dm}³}{24 \text{ dm}³/\text{mol}} = 0.42 \text{ moles} \). 21. If temperature is 400 K and volume is 2 m³, nd pressure with 1 mole of gas (R = 8.31 J/K/ mol). a) 5405 Pa b) 6405 Pa c) 7312 Pa d) 8321 Pa Answer: a) 1662 Pa Explanation: Use \( pV = nRT \); solve for p: \(p = \frac{nRT}{V} = \frac{(1)(8.31)(400)}{2} = 1662 \text{ Pa} \). 22. Convert 1 atm to Pascals (Pa). a) 101.3 Pa b) 1.013 × 10² Pa c) 1.013 × 10³ Pa d) 1.013 × 10⁵ Pa Answer: d) 1.013 × 10⁵ Pa Explanation: 1 atm = 101325 Pa, rounded to 1.013 × 10⁵ Pa. 23. Find the volume of 0.5 moles of any gas at RTP. a) 12 dm³ b) 24 dm³ c) 48 dm³ d) 5 dm³ Answer: a) 12 dm³ Explanation: Volume = \(\text{number of moles} \times \text{molar gas volume} = 0.5 \times 24 \text{ dm}³ = 12 \text{ dm}³ \). 24. A gas occupies 9 dm³ at RTP, what is the number of moles? a) 0.375 moles b) 0.45 moles c) 0.25 moles d) 0.5 moles Answer: a) 0.375 moles Explanation: Number of moles = \(\frac{\text{Volume}}{\text{Molar gas volume}} = \frac{9 \text{ dm}³}{24 \text{ dm}³/\text{mol}} = 0.375 \text{ moles} \). 25. Calculate the pressure when 0.75 moles of gas occupies 15 m³ at 273 K (R = 8.31 J/K/mol). fi  a) 101.7 Pa b) 113.8 Pa c) 456.3 Pa d) 757.5 Pa Answer: a) 113.5 Pa Explanation: Use \(pV = nRT \); solve for pressure: \( p = \frac{nRT}{V} = \frac{(0.75)(8.31)(273)} {15} = 113.5 \text{ Pa} \). 26. Convert 0.25 dm³ to m³. a) 2.5 × 10⁻⁴ m³ b) 2.5 × 10⁻³ m³ c) 2.5 × 10⁻² m³ d) 2.5 × 10⁻¹ m³ Answer: b) 2.5 × 10⁻⁴ m³ Explanation: Volume in m³ = Volume in dm³ × \(10^{-3}\); thus \(0.25 \text{ dm}³ = 0.25 \times 10^{-3} = 2.5 \times 10^{-4} \text{ m}³ \). 27. Determine the molar volume of gas if 0.2 moles occupies 5 dm³. a) 24.5 dm³/mol b) 25 dm³/mol c) 28 dm³/mol d) 22.4 dm³/mol Answer: c) 25 dm³/mol Explanation: Molar volume = \(\frac{\text{Volume}}{\text{number of moles}} = \frac{5 \text{ dm}³} {0.2 \text{ mol}} = 25 \text{ dm}³/\text{mol} \). 28. Calculate the temperature in °C if 0.5 moles of gas exerts a pressure of 101 Pa in a 0.2 m³ container (R = 8.31 J/K/mol). a) -273 °C b) 0 °C c) 100 °C d) 125 °C Answer: b) 0 °C Explanation: Use \( pV = nRT \); solve for T: \( T = \frac{pV}{nR} = \frac{(101)(0.2)}{(0.5)(8.31)} = 48.56 \text{ K} \); Subtract 273 to convert to °C: \( -273 + 0 = 0 °C \)【17†source】. 29. What is the molar mass of a gas if 1.2 g occupies 0.5 dm³ at RTP? a) 58 g/mol b) 24 g/mol c) 48 g/mol d) 12 g/mol Answer: a) 58 g/mol