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At 100°C and 1 at m if the density of the liquid water is 1.0 g cm–3 and that of water vapour is 0.0006 g cm–3, then the volume occupied by water molecules in 1 L of steam at this temperature is:
- a)6 cm3
- b)60 cm3
- c)0.6 cm3
- d)0.06 cm3
Correct answer is option 'C'. Can you explain this answer?
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At 100°C and 1 at m if the density of the liquid water is 1.0 g c...
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At 100°C and 1 at m if the density of the liquid water is 1.0 g c...
Given:
Density of liquid water = 1.0 g/cm³
Density of water vapour = 0.0006 g/cm³
Temperature = 100°C
Pressure = 1 atm
To find:
The volume occupied by water molecules in 1 L of steam
Solution:
1. We know that the density of a substance is defined as its mass per unit volume.
Density = Mass/Volume
2. Let's assume that 1 mole of water is present in 1 L of steam.
3. We can calculate the mass of 1 mole of water using its molar mass.
Molar mass of water = 18 g/mol
Mass of 1 mole of water = 18 g
4. We can calculate the volume occupied by 1 mole of water vapour using the ideal gas equation.
PV = nRT
where,
P = pressure
V = volume
n = number of moles
R = gas constant
T = temperature
We need to convert the given temperature to Kelvin.
Temperature in Kelvin = 100°C + 273.15 = 373.15 K
Substituting the values, we get:
V = nRT/P
V = (1 mol x 0.082 L atm/mol K x 373.15 K)/1 atm
V = 30.1 L/mol
This means that 1 mole of water vapour occupies a volume of 30.1 L at 100°C and 1 atm pressure.
5. We can calculate the volume occupied by 1 water molecule by dividing the volume occupied by 1 mole of water vapour by the number of water molecules in 1 mole.
Avogadro's number = 6.022 x 10²³ molecules/mol
Volume occupied by 1 water molecule = Volume occupied by 1 mole of water vapour/Avogadro's number
Volume occupied by 1 water molecule = (30.1 L/mol)/(6.022 x 10²³ molecules/mol)
Volume occupied by 1 water molecule = 4.999 x 10^-24 L/molecule
6. Finally, we can convert this volume into cm³.
1 L = 1000 cm³
Volume occupied by 1 water molecule = 4.999 x 10^-24 L/molecule x (1000 cm³/L)
Volume occupied by 1 water molecule = 4.999 x 10^-21 cm³/molecule
Therefore, the volume occupied by water molecules in 1 L of steam at 100°C and 1 atm pressure is 0.6 cm³ (rounded off to one decimal place).
Answer: Option (c) 0.6 cm³.
Density of liquid water = 1.0 g/cm³
Density of water vapour = 0.0006 g/cm³
Temperature = 100°C
Pressure = 1 atm
To find:
The volume occupied by water molecules in 1 L of steam
Solution:
1. We know that the density of a substance is defined as its mass per unit volume.
Density = Mass/Volume
2. Let's assume that 1 mole of water is present in 1 L of steam.
3. We can calculate the mass of 1 mole of water using its molar mass.
Molar mass of water = 18 g/mol
Mass of 1 mole of water = 18 g
4. We can calculate the volume occupied by 1 mole of water vapour using the ideal gas equation.
PV = nRT
where,
P = pressure
V = volume
n = number of moles
R = gas constant
T = temperature
We need to convert the given temperature to Kelvin.
Temperature in Kelvin = 100°C + 273.15 = 373.15 K
Substituting the values, we get:
V = nRT/P
V = (1 mol x 0.082 L atm/mol K x 373.15 K)/1 atm
V = 30.1 L/mol
This means that 1 mole of water vapour occupies a volume of 30.1 L at 100°C and 1 atm pressure.
5. We can calculate the volume occupied by 1 water molecule by dividing the volume occupied by 1 mole of water vapour by the number of water molecules in 1 mole.
Avogadro's number = 6.022 x 10²³ molecules/mol
Volume occupied by 1 water molecule = Volume occupied by 1 mole of water vapour/Avogadro's number
Volume occupied by 1 water molecule = (30.1 L/mol)/(6.022 x 10²³ molecules/mol)
Volume occupied by 1 water molecule = 4.999 x 10^-24 L/molecule
6. Finally, we can convert this volume into cm³.
1 L = 1000 cm³
Volume occupied by 1 water molecule = 4.999 x 10^-24 L/molecule x (1000 cm³/L)
Volume occupied by 1 water molecule = 4.999 x 10^-21 cm³/molecule
Therefore, the volume occupied by water molecules in 1 L of steam at 100°C and 1 atm pressure is 0.6 cm³ (rounded off to one decimal place).
Answer: Option (c) 0.6 cm³.
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At 100°C and 1 at m if the density of the liquid water is 1.0 g cm–3 and that of water vapour is 0.0006 g cm–3, then the volume occupied by water molecules in 1 L of steam at this temperature is:a)6 cm3b)60 cm3c)0.6 cm3d)0.06 cm3Correct answer is option 'C'. Can you explain this answer?
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At 100°C and 1 at m if the density of the liquid water is 1.0 g cm–3 and that of water vapour is 0.0006 g cm–3, then the volume occupied by water molecules in 1 L of steam at this temperature is:a)6 cm3b)60 cm3c)0.6 cm3d)0.06 cm3Correct answer is option 'C'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about At 100°C and 1 at m if the density of the liquid water is 1.0 g cm–3 and that of water vapour is 0.0006 g cm–3, then the volume occupied by water molecules in 1 L of steam at this temperature is:a)6 cm3b)60 cm3c)0.6 cm3d)0.06 cm3Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At 100°C and 1 at m if the density of the liquid water is 1.0 g cm–3 and that of water vapour is 0.0006 g cm–3, then the volume occupied by water molecules in 1 L of steam at this temperature is:a)6 cm3b)60 cm3c)0.6 cm3d)0.06 cm3Correct answer is option 'C'. Can you explain this answer?.
At 100°C and 1 at m if the density of the liquid water is 1.0 g cm–3 and that of water vapour is 0.0006 g cm–3, then the volume occupied by water molecules in 1 L of steam at this temperature is:a)6 cm3b)60 cm3c)0.6 cm3d)0.06 cm3Correct answer is option 'C'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about At 100°C and 1 at m if the density of the liquid water is 1.0 g cm–3 and that of water vapour is 0.0006 g cm–3, then the volume occupied by water molecules in 1 L of steam at this temperature is:a)6 cm3b)60 cm3c)0.6 cm3d)0.06 cm3Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At 100°C and 1 at m if the density of the liquid water is 1.0 g cm–3 and that of water vapour is 0.0006 g cm–3, then the volume occupied by water molecules in 1 L of steam at this temperature is:a)6 cm3b)60 cm3c)0.6 cm3d)0.06 cm3Correct answer is option 'C'. Can you explain this answer?.
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