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A solution is obtained by dissolving 12 g of urea (mol.wt. 60) in one litre of water. Another solution is obtained by dissolving 68.4 g of cane sugar (mol.wt. 342) in one litre of water at same temperature. The lowering of vapour pressure in first solution is
- a)same as that of 2nd solution
- b)one-fifth of the 2nd solution
- c)double that of 2nd solution
- d)five times that 2nd solution
Correct answer is option 'A'. Can you explain this answer?
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A solution is obtained by dissolving 12 g of urea (mol.wt. 60) in one ...
Option A is correct
we known that in the first solution number of the moles of urea.
mass of urea 1
= -------------------------- × ---------
m.wt. of urea V
12 1
= ------------------- × ----------- = 0.2
60 1
In second solution the number of moles of cane sugar.
mass of cane sugar
= --------------------------------------
m. wt. of cane sugar
68.4 1
= ------------- × ------- = 0.2
342 1
we known that in the first solution number of the moles of urea.
mass of urea 1
= -------------------------- × ---------
m.wt. of urea V
12 1
= ------------------- × ----------- = 0.2
60 1
In second solution the number of moles of cane sugar.
mass of cane sugar
= --------------------------------------
m. wt. of cane sugar
68.4 1
= ------------- × ------- = 0.2
342 1
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Community Answer
A solution is obtained by dissolving 12 g of urea (mol.wt. 60) in one ...
Understanding the problem:
In this problem, we are given two solutions - one containing urea and the other containing cane sugar. We need to determine the lowering of vapor pressure in the two solutions.
Solution:
The lowering of vapor pressure in a solution is given by Raoult's law, which states that the vapor pressure of a component in a solution is directly proportional to its mole fraction in the solution.
Let's calculate the mole fraction of urea in the first solution:
Molecular weight of urea (CH₄N₂O) = 60 g/mol
Mass of urea = 12 g
Number of moles of urea = Mass of urea / Molecular weight of urea = 12/60 = 0.2 mol
Mole fraction of urea = Moles of urea / Total moles of solute = 0.2 / (0.2 + 1) = 0.2/1.2 = 1/6
Similarly, let's calculate the mole fraction of cane sugar in the second solution:
Molecular weight of cane sugar (C₁₂H₂₂O₁₁) = 342 g/mol
Mass of cane sugar = 68.4 g
Number of moles of cane sugar = Mass of cane sugar / Molecular weight of cane sugar = 68.4/342 = 0.2 mol
Mole fraction of cane sugar = Moles of cane sugar / Total moles of solute = 0.2 / (0.2 + 1) = 0.2/1.2 = 1/6
Conclusion:
The mole fractions of urea and cane sugar in their respective solutions are both 1/6. Since the vapor pressure lowering is directly proportional to the mole fraction, the lowering of vapor pressure in both solutions is the same.
Therefore, the correct answer is option 'A': The lowering of vapor pressure in the first solution is the same as that of the second solution.
In this problem, we are given two solutions - one containing urea and the other containing cane sugar. We need to determine the lowering of vapor pressure in the two solutions.
Solution:
The lowering of vapor pressure in a solution is given by Raoult's law, which states that the vapor pressure of a component in a solution is directly proportional to its mole fraction in the solution.
Let's calculate the mole fraction of urea in the first solution:
Molecular weight of urea (CH₄N₂O) = 60 g/mol
Mass of urea = 12 g
Number of moles of urea = Mass of urea / Molecular weight of urea = 12/60 = 0.2 mol
Mole fraction of urea = Moles of urea / Total moles of solute = 0.2 / (0.2 + 1) = 0.2/1.2 = 1/6
Similarly, let's calculate the mole fraction of cane sugar in the second solution:
Molecular weight of cane sugar (C₁₂H₂₂O₁₁) = 342 g/mol
Mass of cane sugar = 68.4 g
Number of moles of cane sugar = Mass of cane sugar / Molecular weight of cane sugar = 68.4/342 = 0.2 mol
Mole fraction of cane sugar = Moles of cane sugar / Total moles of solute = 0.2 / (0.2 + 1) = 0.2/1.2 = 1/6
Conclusion:
The mole fractions of urea and cane sugar in their respective solutions are both 1/6. Since the vapor pressure lowering is directly proportional to the mole fraction, the lowering of vapor pressure in both solutions is the same.
Therefore, the correct answer is option 'A': The lowering of vapor pressure in the first solution is the same as that of the second solution.
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A solution is obtained by dissolving 12 g of urea (mol.wt. 60) in one litre of water. Another solution is obtained by dissolving 68.4 g of cane sugar (mol.wt. 342) in one litre of water at same temperature. The lowering of vapour pressure in first solution isa)same as that of 2nd solutionb)one-fifth of the 2nd solutionc)double that of 2nd solutiond)five times that 2nd solutionCorrect answer is option 'A'. Can you explain this answer?
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A solution is obtained by dissolving 12 g of urea (mol.wt. 60) in one litre of water. Another solution is obtained by dissolving 68.4 g of cane sugar (mol.wt. 342) in one litre of water at same temperature. The lowering of vapour pressure in first solution isa)same as that of 2nd solutionb)one-fifth of the 2nd solutionc)double that of 2nd solutiond)five times that 2nd solutionCorrect answer is option 'A'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A solution is obtained by dissolving 12 g of urea (mol.wt. 60) in one litre of water. Another solution is obtained by dissolving 68.4 g of cane sugar (mol.wt. 342) in one litre of water at same temperature. The lowering of vapour pressure in first solution isa)same as that of 2nd solutionb)one-fifth of the 2nd solutionc)double that of 2nd solutiond)five times that 2nd solutionCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A solution is obtained by dissolving 12 g of urea (mol.wt. 60) in one litre of water. Another solution is obtained by dissolving 68.4 g of cane sugar (mol.wt. 342) in one litre of water at same temperature. The lowering of vapour pressure in first solution isa)same as that of 2nd solutionb)one-fifth of the 2nd solutionc)double that of 2nd solutiond)five times that 2nd solutionCorrect answer is option 'A'. Can you explain this answer?.
A solution is obtained by dissolving 12 g of urea (mol.wt. 60) in one litre of water. Another solution is obtained by dissolving 68.4 g of cane sugar (mol.wt. 342) in one litre of water at same temperature. The lowering of vapour pressure in first solution isa)same as that of 2nd solutionb)one-fifth of the 2nd solutionc)double that of 2nd solutiond)five times that 2nd solutionCorrect answer is option 'A'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A solution is obtained by dissolving 12 g of urea (mol.wt. 60) in one litre of water. Another solution is obtained by dissolving 68.4 g of cane sugar (mol.wt. 342) in one litre of water at same temperature. The lowering of vapour pressure in first solution isa)same as that of 2nd solutionb)one-fifth of the 2nd solutionc)double that of 2nd solutiond)five times that 2nd solutionCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A solution is obtained by dissolving 12 g of urea (mol.wt. 60) in one litre of water. Another solution is obtained by dissolving 68.4 g of cane sugar (mol.wt. 342) in one litre of water at same temperature. The lowering of vapour pressure in first solution isa)same as that of 2nd solutionb)one-fifth of the 2nd solutionc)double that of 2nd solutiond)five times that 2nd solutionCorrect answer is option 'A'. Can you explain this answer?.
Solutions for A solution is obtained by dissolving 12 g of urea (mol.wt. 60) in one litre of water. Another solution is obtained by dissolving 68.4 g of cane sugar (mol.wt. 342) in one litre of water at same temperature. The lowering of vapour pressure in first solution isa)same as that of 2nd solutionb)one-fifth of the 2nd solutionc)double that of 2nd solutiond)five times that 2nd solutionCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for NEET.
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