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One molal aqueous glucose solution has density 1 .180 gm L-1. Also, Kf / Kb = 3 . Thus,
- a)Δ T/Δ Tf=3
- b)osmotic pressure at 300 K is 24.63 atm
- c)osmotic pressure at 300 K is 29.06 atm
- d)Δ T/Δ Tf=0.33
Correct answer is option 'B,D'. Can you explain this answer?
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Verified Answer
One molal aqueous glucose solution has density 1 .180 gm L-1. Also, Kf...
Glucose solution is = 1 mol kg-1 water
Solute (glucose) = 180g
Solvent (water) = 1000 g
Most Upvoted Answer
One molal aqueous glucose solution has density 1 .180 gm L-1. Also, Kf...
To solve this problem, we need to use the concepts of molality, density, freezing point depression, and osmotic pressure.
1. Calculation of molality (m):
Molality (m) is defined as the number of moles of solute per kilogram of solvent. Since we have a one molal aqueous glucose solution, it means that there is one mole of glucose dissolved in one kilogram of water.
2. Calculation of moles of glucose:
To find the number of moles of glucose, we need to know the density of the solution. The density is given as 1.180 gm L-1, which means that one liter of the solution has a mass of 1.180 grams. Since the density is mass per unit volume, we can calculate the mass of one kilogram of the solution.
Mass of 1 L of solution = 1.180 g
Mass of 1000 mL of solution = 1.180 g
Mass of 1000 g (1 kg) of solution = 1.180 g
Therefore, the mass of 1 kg of the solution is 1.180 g.
3. Calculation of moles of solvent (water):
Since the molality is defined as the number of moles of solute per kilogram of solvent, we need to find the number of moles of water in one kilogram of the solution. The molar mass of water is 18.015 g/mol.
Number of moles of water = Mass of water / molar mass of water
Number of moles of water = 1000 g / 18.015 g/mol
Number of moles of water = 55.51 mol
4. Calculation of moles of glucose:
Since the molality is one molal, it means that there is one mole of glucose in one kilogram of water.
Number of moles of glucose = 1 mol
5. Calculation of freezing point depression:
The freezing point depression (ΔTf) is given by the equation:
ΔTf = Kf * m
Here, Kf is the cryoscopic constant, which is the molal freezing point depression constant, and m is the molality.
Since Kf / Kb = 3, we can calculate the value of Kf as:
Kf = 3 * Kb
6. Calculation of osmotic pressure:
The osmotic pressure (π) can be calculated using the equation:
π = i * M * R * T
Here, i is the van't Hoff factor, which is the number of particles the solute dissociates into in the solution. For glucose, i = 1.
M is the molarity of the solution, which can be calculated using the equation:
M = m * (density of water / molar mass of water)
R is the ideal gas constant (0.0821 L * atm / (mol * K)), and T is the temperature in Kelvin.
For 300 K, we can calculate the osmotic pressure using the given equation.
7. Calculation of T/Tf:
T/Tf is the ratio of the actual temperature to the freezing point temperature. Since we know the value of ΔTf from step 5, we can calculate the freezing point temperature (Tf) using the equation:
Tf = T - ΔTf
For part (a), T/T
1. Calculation of molality (m):
Molality (m) is defined as the number of moles of solute per kilogram of solvent. Since we have a one molal aqueous glucose solution, it means that there is one mole of glucose dissolved in one kilogram of water.
2. Calculation of moles of glucose:
To find the number of moles of glucose, we need to know the density of the solution. The density is given as 1.180 gm L-1, which means that one liter of the solution has a mass of 1.180 grams. Since the density is mass per unit volume, we can calculate the mass of one kilogram of the solution.
Mass of 1 L of solution = 1.180 g
Mass of 1000 mL of solution = 1.180 g
Mass of 1000 g (1 kg) of solution = 1.180 g
Therefore, the mass of 1 kg of the solution is 1.180 g.
3. Calculation of moles of solvent (water):
Since the molality is defined as the number of moles of solute per kilogram of solvent, we need to find the number of moles of water in one kilogram of the solution. The molar mass of water is 18.015 g/mol.
Number of moles of water = Mass of water / molar mass of water
Number of moles of water = 1000 g / 18.015 g/mol
Number of moles of water = 55.51 mol
4. Calculation of moles of glucose:
Since the molality is one molal, it means that there is one mole of glucose in one kilogram of water.
Number of moles of glucose = 1 mol
5. Calculation of freezing point depression:
The freezing point depression (ΔTf) is given by the equation:
ΔTf = Kf * m
Here, Kf is the cryoscopic constant, which is the molal freezing point depression constant, and m is the molality.
Since Kf / Kb = 3, we can calculate the value of Kf as:
Kf = 3 * Kb
6. Calculation of osmotic pressure:
The osmotic pressure (π) can be calculated using the equation:
π = i * M * R * T
Here, i is the van't Hoff factor, which is the number of particles the solute dissociates into in the solution. For glucose, i = 1.
M is the molarity of the solution, which can be calculated using the equation:
M = m * (density of water / molar mass of water)
R is the ideal gas constant (0.0821 L * atm / (mol * K)), and T is the temperature in Kelvin.
For 300 K, we can calculate the osmotic pressure using the given equation.
7. Calculation of T/Tf:
T/Tf is the ratio of the actual temperature to the freezing point temperature. Since we know the value of ΔTf from step 5, we can calculate the freezing point temperature (Tf) using the equation:
Tf = T - ΔTf
For part (a), T/T
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One molal aqueous glucose solution has density 1 .180 gm L-1. Also, Kf / Kb = 3 . Thus,a)T/Tf=3b)osmotic pressure at 300 K is 24.63 atmc)osmotic pressure at 300 K is 29.06 atmd)T/Tf=0.33Correct answer is option 'B,D'. Can you explain this answer?
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One molal aqueous glucose solution has density 1 .180 gm L-1. Also, Kf / Kb = 3 . Thus,a)T/Tf=3b)osmotic pressure at 300 K is 24.63 atmc)osmotic pressure at 300 K is 29.06 atmd)T/Tf=0.33Correct answer is option 'B,D'. 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 One molal aqueous glucose solution has density 1 .180 gm L-1. Also, Kf / Kb = 3 . Thus,a)T/Tf=3b)osmotic pressure at 300 K is 24.63 atmc)osmotic pressure at 300 K is 29.06 atmd)T/Tf=0.33Correct answer is option 'B,D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One molal aqueous glucose solution has density 1 .180 gm L-1. Also, Kf / Kb = 3 . Thus,a)T/Tf=3b)osmotic pressure at 300 K is 24.63 atmc)osmotic pressure at 300 K is 29.06 atmd)T/Tf=0.33Correct answer is option 'B,D'. Can you explain this answer?.
One molal aqueous glucose solution has density 1 .180 gm L-1. Also, Kf / Kb = 3 . Thus,a)T/Tf=3b)osmotic pressure at 300 K is 24.63 atmc)osmotic pressure at 300 K is 29.06 atmd)T/Tf=0.33Correct answer is option 'B,D'. 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 One molal aqueous glucose solution has density 1 .180 gm L-1. Also, Kf / Kb = 3 . Thus,a)T/Tf=3b)osmotic pressure at 300 K is 24.63 atmc)osmotic pressure at 300 K is 29.06 atmd)T/Tf=0.33Correct answer is option 'B,D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One molal aqueous glucose solution has density 1 .180 gm L-1. Also, Kf / Kb = 3 . Thus,a)T/Tf=3b)osmotic pressure at 300 K is 24.63 atmc)osmotic pressure at 300 K is 29.06 atmd)T/Tf=0.33Correct answer is option 'B,D'. Can you explain this answer?.
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