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18 gm of glucose is added to 178.2 gm water . the vapour pressure of water (in torr) for this aqueous solution at 373 k is?
Most Upvoted Answer
18 gm of glucose is added to 178.2 gm water . the vapour pressure of w...
Calculation of Vapour Pressure of Water in Aqueous Solution
Given:
Mass of glucose (solute) = 18 gm
Mass of water (solvent) = 178.2 gm
Temperature (T) = 373 K
To find:
Vapour pressure of water (P)
Step 1: Calculate the mole fraction of water (Xw)
Mole fraction of water (Xw) = Number of moles of water / Total number of moles
Number of moles of water = Mass of water / Molar mass of water
Molar mass of water = 18 g/mol
Number of moles of water = 178.2 / 18 = 9.9 mol
Number of moles of glucose = Mass of glucose / Molar mass of glucose
Molar mass of glucose = 180 g/mol
Number of moles of glucose = 18 / 180 = 0.1 mol
Total number of moles = 9.9 + 0.1 = 10 mol
Xw = 9.9 / 10 = 0.99
Step 2: Calculate the vapour pressure of pure water (Po)
At 373 K, the vapour pressure of pure water is 760 torr.
Step 3: Calculate the vapour pressure of water in the solution (P)
P = Xw * Po
P = 0.99 * 760
P = 752.4 torr
Answer:
The vapour pressure of water in the aqueous solution is 752.4 torr at 373 K.
Explanation:
When a solute is added to a solvent, the vapour pressure of the solvent decreases. This is due to the fact that the solute molecules occupy some of the space at the surface of the solvent. As a result, fewer solvent molecules can escape into the vapour phase, leading to a decrease in vapour pressure. The magnitude of this decrease depends on the concentration of the solute in the solution. The mole fraction of water in the solution is calculated to determine the concentration of water. The vapour pressure of pure water is then used to determine the vapour pressure of water in the solution.
Given:
Mass of glucose (solute) = 18 gm
Mass of water (solvent) = 178.2 gm
Temperature (T) = 373 K
To find:
Vapour pressure of water (P)
Step 1: Calculate the mole fraction of water (Xw)
Mole fraction of water (Xw) = Number of moles of water / Total number of moles
Number of moles of water = Mass of water / Molar mass of water
Molar mass of water = 18 g/mol
Number of moles of water = 178.2 / 18 = 9.9 mol
Number of moles of glucose = Mass of glucose / Molar mass of glucose
Molar mass of glucose = 180 g/mol
Number of moles of glucose = 18 / 180 = 0.1 mol
Total number of moles = 9.9 + 0.1 = 10 mol
Xw = 9.9 / 10 = 0.99
Step 2: Calculate the vapour pressure of pure water (Po)
At 373 K, the vapour pressure of pure water is 760 torr.
Step 3: Calculate the vapour pressure of water in the solution (P)
P = Xw * Po
P = 0.99 * 760
P = 752.4 torr
Answer:
The vapour pressure of water in the aqueous solution is 752.4 torr at 373 K.
Explanation:
When a solute is added to a solvent, the vapour pressure of the solvent decreases. This is due to the fact that the solute molecules occupy some of the space at the surface of the solvent. As a result, fewer solvent molecules can escape into the vapour phase, leading to a decrease in vapour pressure. The magnitude of this decrease depends on the concentration of the solute in the solution. The mole fraction of water in the solution is calculated to determine the concentration of water. The vapour pressure of pure water is then used to determine the vapour pressure of water in the solution.
Community Answer
18 gm of glucose is added to 178.2 gm water . the vapour pressure of w...
Mole fraction of water =9.9/10=0.99and V. pressure of pure water at 373K is 760torr...so,V.p. =760× 0.99=752.4
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18 gm of glucose is added to 178.2 gm water . the vapour pressure of water (in torr) for this aqueous solution at 373 k is?
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18 gm of glucose is added to 178.2 gm water . the vapour pressure of water (in torr) for this aqueous solution at 373 k is? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 18 gm of glucose is added to 178.2 gm water . the vapour pressure of water (in torr) for this aqueous solution at 373 k is? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 18 gm of glucose is added to 178.2 gm water . the vapour pressure of water (in torr) for this aqueous solution at 373 k is?.
18 gm of glucose is added to 178.2 gm water . the vapour pressure of water (in torr) for this aqueous solution at 373 k is? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 18 gm of glucose is added to 178.2 gm water . the vapour pressure of water (in torr) for this aqueous solution at 373 k is? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 18 gm of glucose is added to 178.2 gm water . the vapour pressure of water (in torr) for this aqueous solution at 373 k is?.
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