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In which case is the no.of molecules of water maximum? A) 18ml of water B) 0.18g of water C)0.00224 l of water vapour at 1 atm and 273k D) 10^-3mol of water?
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In which case is the no.of molecules of water maximum? A) 18ml of wate...
Answer:
The no. of molecules of water is maximum in option D) 10^-3mol of water.
Explanation:
To calculate the no. of molecules of water, we need to use Avogadro's number. Avogadro's number is defined as the number of atoms or molecules contained in one mole of substance and is equal to 6.022 x 10^23.
Let's calculate the no. of molecules of water in each option.
A) 18ml of water:
We need to convert ml to moles using the density of water, which is 1g/ml.
18ml of water = 18g of water
No. of moles of water = 18g / 18 g/mol = 1 mole
No. of molecules of water = 1 mole x Avogadro's number = 6.022 x 10^23
B) 0.18g of water:
No. of moles of water = 0.18g / 18 g/mol = 0.01 mole
No. of molecules of water = 0.01 mole x Avogadro's number = 6.022 x 10^21
C) 0.00224 l of water vapour at 1 atm and 273k:
No. of moles of water = (0.00224 L x 1 atm) / (0.0821 L atm/mol K x 273K) = 0.0001 mole
No. of molecules of water = 0.0001 mole x Avogadro's number = 6.022 x 10^19
D) 10^-3mol of water:
No. of molecules of water = 10^-3 mole x Avogadro's number = 6.022 x 10^20
Therefore, the no. of molecules of water is maximum in option D) 10^-3mol of water.
The no. of molecules of water is maximum in option D) 10^-3mol of water.
Explanation:
To calculate the no. of molecules of water, we need to use Avogadro's number. Avogadro's number is defined as the number of atoms or molecules contained in one mole of substance and is equal to 6.022 x 10^23.
Let's calculate the no. of molecules of water in each option.
A) 18ml of water:
We need to convert ml to moles using the density of water, which is 1g/ml.
18ml of water = 18g of water
No. of moles of water = 18g / 18 g/mol = 1 mole
No. of molecules of water = 1 mole x Avogadro's number = 6.022 x 10^23
B) 0.18g of water:
No. of moles of water = 0.18g / 18 g/mol = 0.01 mole
No. of molecules of water = 0.01 mole x Avogadro's number = 6.022 x 10^21
C) 0.00224 l of water vapour at 1 atm and 273k:
No. of moles of water = (0.00224 L x 1 atm) / (0.0821 L atm/mol K x 273K) = 0.0001 mole
No. of molecules of water = 0.0001 mole x Avogadro's number = 6.022 x 10^19
D) 10^-3mol of water:
No. of molecules of water = 10^-3 mole x Avogadro's number = 6.022 x 10^20
Therefore, the no. of molecules of water is maximum in option D) 10^-3mol of water.
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In which case is the no.of molecules of water maximum? A) 18ml of wate...
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In which case is the no.of molecules of water maximum? A) 18ml of water B) 0.18g of water C)0.00224 l of water vapour at 1 atm and 273k D) 10^-3mol of water? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about In which case is the no.of molecules of water maximum? A) 18ml of water B) 0.18g of water C)0.00224 l of water vapour at 1 atm and 273k D) 10^-3mol of water? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In which case is the no.of molecules of water maximum? A) 18ml of water B) 0.18g of water C)0.00224 l of water vapour at 1 atm and 273k D) 10^-3mol of water?.
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