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6.0 g of urea was dissolved in 9.9 moles of water. If the vapour pressure of pure water is P°, the vapour pressure of solution is:
- a)0.10 P°
- b)1.10 P°
- c)0.90 P°
- d)0.99 P°
Correct answer is option 'D'. Can you explain this answer?
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6.0 g of urea was dissolved in 9.9 moles of water. If the vapour press...
The answer is c.
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6.0 g of urea was dissolved in 9.9 moles of water. If the vapour press...
Solution:
In this question, we have to find the vapour pressure of the solution which is formed by dissolving urea in water.
The vapour pressure of water decreases when a non-volatile solute is added to it. This is because the non-volatile solute lowers the total number of water molecules at the surface of the solution, which in turn reduces the number of water molecules that can escape from the solution to form vapour.
The decrease in vapour pressure is given by Raoult's law, which states that the partial vapour pressure of each component in a solution is proportional to its mole fraction in the solution.
Let us first calculate the mole fraction of water in the solution:
Molar mass of urea = 60 g/mol
Molar mass of water = 18 g/mol
Number of moles of urea = 6.0/60 = 0.1 mol
Number of moles of water = 9.9 mol
Total number of moles in the solution = 0.1 + 9.9 = 10.0 mol
Mole fraction of water = 9.9/10.0 = 0.99
Now, according to Raoult's law, the vapour pressure of the solution (P solution) is given by:
P solution = X water * P
where X water is the mole fraction of water in the solution and P is the vapour pressure of pure water.
Substituting the values, we get:
P solution = 0.99 * P = 0.99P
Therefore, the vapour pressure of the solution is 0.99 times the vapour pressure of pure water.
Hence, the correct option is D) 0.99 P.
In this question, we have to find the vapour pressure of the solution which is formed by dissolving urea in water.
The vapour pressure of water decreases when a non-volatile solute is added to it. This is because the non-volatile solute lowers the total number of water molecules at the surface of the solution, which in turn reduces the number of water molecules that can escape from the solution to form vapour.
The decrease in vapour pressure is given by Raoult's law, which states that the partial vapour pressure of each component in a solution is proportional to its mole fraction in the solution.
Let us first calculate the mole fraction of water in the solution:
Molar mass of urea = 60 g/mol
Molar mass of water = 18 g/mol
Number of moles of urea = 6.0/60 = 0.1 mol
Number of moles of water = 9.9 mol
Total number of moles in the solution = 0.1 + 9.9 = 10.0 mol
Mole fraction of water = 9.9/10.0 = 0.99
Now, according to Raoult's law, the vapour pressure of the solution (P solution) is given by:
P solution = X water * P
where X water is the mole fraction of water in the solution and P is the vapour pressure of pure water.
Substituting the values, we get:
P solution = 0.99 * P = 0.99P
Therefore, the vapour pressure of the solution is 0.99 times the vapour pressure of pure water.
Hence, the correct option is D) 0.99 P.
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6.0 g of urea was dissolved in 9.9 moles of water. If the vapour pressure of pure water is P°, the vapour pressure of solution is:a)0.10 P°b)1.10 P°c)0.90 P°d)0.99 P°Correct answer is option 'D'. Can you explain this answer?
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6.0 g of urea was dissolved in 9.9 moles of water. If the vapour pressure of pure water is P°, the vapour pressure of solution is:a)0.10 P°b)1.10 P°c)0.90 P°d)0.99 P°Correct answer is option 'D'. 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 6.0 g of urea was dissolved in 9.9 moles of water. If the vapour pressure of pure water is P°, the vapour pressure of solution is:a)0.10 P°b)1.10 P°c)0.90 P°d)0.99 P°Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 6.0 g of urea was dissolved in 9.9 moles of water. If the vapour pressure of pure water is P°, the vapour pressure of solution is:a)0.10 P°b)1.10 P°c)0.90 P°d)0.99 P°Correct answer is option 'D'. Can you explain this answer?.
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