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Volume occupied by one molecule of water (density = 1 g cm–3) is : [2008]
- a)9.0 × 10–23 cm3
- b)6.023 × 10– 23 cm3
- c)3.0 × 10–23 cm3
- d)5.5 × 10– 23 cm3
Correct answer is option 'C'. Can you explain this answer?
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$^-3$) at room temperature (25°C) and atmospheric pressure (1 atm).
At room temperature and atmospheric pressure, one molecule of water occupies a volume of approximately 18.1 cubic angstroms (Å$^3$). This can be calculated using the ideal gas law, which states that PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the gas constant, and T is temperature. Assuming one molecule of water is equivalent to one mole of gas, and substituting the values for pressure, temperature, and the gas constant, we can solve for volume:
V = nRT/P = (1 mol x 0.08206 L atm mol$^-1$ K$^-1$ x 298 K) / (1 atm) = 24.465 L mol$^-1$
Converting liters to cubic angstroms and dividing by Avogadro's number (6.022 x 10$^{23}$) to get the volume per molecule, we get:
V = 24.465 x 10$^24$ Å$^3$ mol$^-1$ / (6.022 x 10$^{23}$) = 18.1 Å$^3$ per molecule
At room temperature and atmospheric pressure, one molecule of water occupies a volume of approximately 18.1 cubic angstroms (Å$^3$). This can be calculated using the ideal gas law, which states that PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the gas constant, and T is temperature. Assuming one molecule of water is equivalent to one mole of gas, and substituting the values for pressure, temperature, and the gas constant, we can solve for volume:
V = nRT/P = (1 mol x 0.08206 L atm mol$^-1$ K$^-1$ x 298 K) / (1 atm) = 24.465 L mol$^-1$
Converting liters to cubic angstroms and dividing by Avogadro's number (6.022 x 10$^{23}$) to get the volume per molecule, we get:
V = 24.465 x 10$^24$ Å$^3$ mol$^-1$ / (6.022 x 10$^{23}$) = 18.1 Å$^3$ per molecule
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Volume occupied by one molecule of water (density = 1 g cm–3) is : [2008]a)9.0 × 10–23 cm3b)6.023 × 10– 23 cm3c)3.0 × 10–23 cm3d)5.5 × 10– 23 cm3Correct answer is option 'C'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Volume occupied by one molecule of water (density = 1 g cm–3) is : [2008]a)9.0 × 10–23 cm3b)6.023 × 10– 23 cm3c)3.0 × 10–23 cm3d)5.5 × 10– 23 cm3Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Volume occupied by one molecule of water (density = 1 g cm–3) is : [2008]a)9.0 × 10–23 cm3b)6.023 × 10– 23 cm3c)3.0 × 10–23 cm3d)5.5 × 10– 23 cm3Correct answer is option 'C'. Can you explain this answer?.
Volume occupied by one molecule of water (density = 1 g cm–3) is : [2008]a)9.0 × 10–23 cm3b)6.023 × 10– 23 cm3c)3.0 × 10–23 cm3d)5.5 × 10– 23 cm3Correct answer is option 'C'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Volume occupied by one molecule of water (density = 1 g cm–3) is : [2008]a)9.0 × 10–23 cm3b)6.023 × 10– 23 cm3c)3.0 × 10–23 cm3d)5.5 × 10– 23 cm3Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Volume occupied by one molecule of water (density = 1 g cm–3) is : [2008]a)9.0 × 10–23 cm3b)6.023 × 10– 23 cm3c)3.0 × 10–23 cm3d)5.5 × 10– 23 cm3Correct answer is option 'C'. Can you explain this answer?.
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Volume occupied by one molecule of water (density = 1 g cm–3) is : [2008]a)9.0 × 10–23 cm3b)6.023 × 10– 23 cm3c)3.0 × 10–23 cm3d)5.5 × 10– 23 cm3Correct answer is option 'C'. Can you explain this answer?, a detailed solution for Volume occupied by one molecule of water (density = 1 g cm–3) is : [2008]a)9.0 × 10–23 cm3b)6.023 × 10– 23 cm3c)3.0 × 10–23 cm3d)5.5 × 10– 23 cm3Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of Volume occupied by one molecule of water (density = 1 g cm–3) is : [2008]a)9.0 × 10–23 cm3b)6.023 × 10– 23 cm3c)3.0 × 10–23 cm3d)5.5 × 10– 23 cm3Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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