The Gaseous And Liquid State — Practice Quiz
NIOS · Class 12 · Chemistry
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Quick Quiz: The Gaseous And Liquid State
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A gas occupies 500 mL at a pressure of 0.20 atm. What pressure is required to compress it to 10 mL at constant temperature?
At constant pressure, a gas occupies 2 L at 1000°C. What will be its volume at 0°C?
Which of the following graphs correctly represents Boyle's Law for a fixed amount of gas at constant temperature?
10 mol of an ideal gas is confined in a container of volume 224 L at 273 K. What is the pressure of the gas? (R = 0.0821 L atm mol⁻¹ K⁻¹)
Sample Questions
A mixture of gases contains 2.74 mol of N₂ and 0.725 mol of O₂ at a total pressure of 1 atm. What is the partial pressure of O₂?
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0.209 atm
Step 1: Apply Dalton's Law of Partial Pressures: pA = XA × pTotal. Step 2: Total moles = 2.74 + 0.725 = 3.465 mol. Step 3: Mole fraction of O₂ = 0.725 / 3.465 = 0.209. Step 4: Partial pressure of O₂ = 0.209 × 1 atm = 0.209 atm. Option B (0.791 atm) is the partial pressure of N₂, not O₂ – a common mix-up. Option C uses number of moles directly as pressure. Option D incorrectly assumes equal mole fractions.
Which of the following correctly explains why NH₃ has an abnormally high boiling point compared to PH₃, AsH₃, and SbH₃?
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NH₃ molecules form intermolecular hydrogen bonds due to the high electronegativity of N
Step 1: In group 15 hydrides, boiling points should increase with molar mass going down the group (PH₃ → SbH₃). NH₃ breaks this trend. Step 2: Nitrogen is highly electronegative (3.0) and has a lone pair. The N–H bond is highly polar, allowing N…H–N hydrogen bonding between molecules. Step 3: Hydrogen bonding is much stronger than ordinary van der Waals forces, requiring more energy (higher temperature) to break – hence higher boiling point. Step 4: Option B is wrong because NH₃ has LOWER molar mass (17) than PH₃ (34), yet boils higher. Option C is wrong because London forces increase with mas
The rates of diffusion of CO₂ and an unknown gas X were found to be in the ratio 1 : 0.5. What is the molar mass of gas X? (Molar mass of CO₂ = 44 g/mol)
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176 g/mol
Step 1: Graham's Law states: rate_A / rate_B = √(M_B / M_A). Step 2: rate_CO₂ / rate_X = √(M_X / M_CO₂) → 1/0.5 = √(M_X / 44). Step 3: 2 = √(M_X / 44) → squaring both sides: 4 = M_X / 44. Step 4: M_X = 4 × 44 = 176 g/mol. Option B (11 g/mol) comes from dividing instead of multiplying. Option C (88 g/mol) results from squaring 2 and multiplying by 22 (using half of M_CO₂). Option D comes from not squaring the ratio at all.
For a real gas, the van der Waals equation is written as (p + an²/V²)(V − nb) = nRT. What does the term 'an²/V²' correct for?
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Intermolecular attractive forces between gas molecules
Step 1: Ideal gas assumes no intermolecular attractive forces. Real gases have attraction between molecules, especially at high pressure. Step 2: Due to these attractive forces, molecules are pulled back from the walls, reducing the actual pressure exerted on the walls compared to ideal pressure. Step 3: So: p_ideal = p_real + an²/V². The term an²/V² is the pressure correction that accounts for these attractive forces ('a' is van der Waals constant related to attraction). Step 4: The term 'nb' corrects for the finite volume of molecules (Option B is the explanation for 'nb', not 'an²/V²'). Opt
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