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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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Most Upvoted Answer
A solution is obtained by dissolving 12 g of urea (mol.wt. 60) in one ...
Solution:
The lowering of vapour pressure is given by the formula
ΔP = P° - P
where,
P° = Vapour pressure of solvent in pure state
P = Vapour pressure of solvent in solution
Let us first calculate the mole fraction of urea and cane sugar in their respective solutions.
Mole fraction of urea = (Number of moles of urea) / (Total number of moles)
Number of moles of urea = (Given mass of urea) / (Molar mass of urea)
= 12 g / 60 g/mol
= 0.2 mol
Total number of moles = (Mass of solute) / (Molar mass of solute) + (Mass of solvent) / (Molar mass of solvent)
= 12 g / 60 g/mol + 1000 g / 18 g/mol
= 0.2 mol + 55.56 mol
= 55.76 mol
Mole fraction of urea = 0.2 / 55.76
= 0.0036
Similarly, we can calculate the mole fraction of cane sugar in its solution.
Mole fraction of cane sugar = (Number of moles of cane sugar) / (Total number of moles)
Number of moles of cane sugar = (Given mass of cane sugar) / (Molar mass of cane sugar)
= 68.4 g / 342 g/mol
= 0.2 mol
Total number of moles = (Mass of solute) / (Molar mass of solute) + (Mass of solvent) / (Molar mass of solvent)
= 68.4 g / 342 g/mol + 1000 g / 18 g/mol
= 0.2 mol + 55.56 mol
= 55.76 mol
Mole fraction of cane sugar = 0.2 / 55.76
= 0.0036
As we can see, the mole fraction of both solutes is the same in their respective solutions.
Now, let us calculate the lowering of vapour pressure in each solution using the formula mentioned above.
For urea solution,
P° = Vapour pressure of water in pure state at the given temperature
= 23.8 mmHg (at 25°C)
P = Vapour pressure of water in urea solution
= Xwater x P°
= (1 - Xurea) x P°
= (1 - 0.0036) x 23.8 mmHg
= 23.72 mmHg
ΔP = P° - P
= 23.8 mmHg - 23.72 mmHg
= 0.08 mmHg
For cane sugar solution,
P° = Vapour pressure of water in pure state at the given temperature
= 23.8 mmHg (at 25°C)
P = Vapour pressure of water in cane sugar solution
= Xwater x P°
= (1 - Xcane sugar) x P°
= (1 - 0.0036) x 23.8 mmHg
= 23.72 mmHg
ΔP = P° - P
= 23.8 mmHg -
The lowering of vapour pressure is given by the formula
ΔP = P° - P
where,
P° = Vapour pressure of solvent in pure state
P = Vapour pressure of solvent in solution
Let us first calculate the mole fraction of urea and cane sugar in their respective solutions.
Mole fraction of urea = (Number of moles of urea) / (Total number of moles)
Number of moles of urea = (Given mass of urea) / (Molar mass of urea)
= 12 g / 60 g/mol
= 0.2 mol
Total number of moles = (Mass of solute) / (Molar mass of solute) + (Mass of solvent) / (Molar mass of solvent)
= 12 g / 60 g/mol + 1000 g / 18 g/mol
= 0.2 mol + 55.56 mol
= 55.76 mol
Mole fraction of urea = 0.2 / 55.76
= 0.0036
Similarly, we can calculate the mole fraction of cane sugar in its solution.
Mole fraction of cane sugar = (Number of moles of cane sugar) / (Total number of moles)
Number of moles of cane sugar = (Given mass of cane sugar) / (Molar mass of cane sugar)
= 68.4 g / 342 g/mol
= 0.2 mol
Total number of moles = (Mass of solute) / (Molar mass of solute) + (Mass of solvent) / (Molar mass of solvent)
= 68.4 g / 342 g/mol + 1000 g / 18 g/mol
= 0.2 mol + 55.56 mol
= 55.76 mol
Mole fraction of cane sugar = 0.2 / 55.76
= 0.0036
As we can see, the mole fraction of both solutes is the same in their respective solutions.
Now, let us calculate the lowering of vapour pressure in each solution using the formula mentioned above.
For urea solution,
P° = Vapour pressure of water in pure state at the given temperature
= 23.8 mmHg (at 25°C)
P = Vapour pressure of water in urea solution
= Xwater x P°
= (1 - Xurea) x P°
= (1 - 0.0036) x 23.8 mmHg
= 23.72 mmHg
ΔP = P° - P
= 23.8 mmHg - 23.72 mmHg
= 0.08 mmHg
For cane sugar solution,
P° = Vapour pressure of water in pure state at the given temperature
= 23.8 mmHg (at 25°C)
P = Vapour pressure of water in cane sugar solution
= Xwater x P°
= (1 - Xcane sugar) x P°
= (1 - 0.0036) x 23.8 mmHg
= 23.72 mmHg
ΔP = P° - P
= 23.8 mmHg -
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A solution is obtained by dissolving 12 g of urea (mol.wt. 60) in one ...
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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 JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE 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 JEE 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 JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE 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 JEE 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?.
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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?, a detailed solution 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? has been provided alongside types of 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? theory, EduRev gives you an
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