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1.00 g of a non-electrolyte solute (molar mass 250 g mol–1) was dissolved in 51.2 g of benzene. If the freezing point depression constant, Kf of benzene is 5.12 K kg mol–1, the freezing point of benzene will be lowered by [2006]
- a)0.3 K
- b)0.5 K
- c)0.4 K
- d)0.2
Correct answer is option 'C'. Can you explain this answer?
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$^{-1}$) is dissolved in 1000 mL of water at 25°C. What is the osmotic pressure of the resulting solution?
First, we need to calculate the molarity of the solution:
Molarity = moles of solute / volume of solution in liters
Since we have 1.00 g of solute and its molar mass is 250 g mol$^{-1}$, we can calculate the number of moles:
moles of solute = 1.00 g / 250 g mol$^{-1}$ = 0.004 mol
The volume of the solution is given as 1000 mL, which is equal to 1.00 L. Therefore, the molarity of the solution is:
Molarity = 0.004 mol / 1.00 L = 0.004 M
Next, we can use the van't Hoff equation to calculate the osmotic pressure:
π = MRT
where π is the osmotic pressure, M is the molarity of the solution, R is the gas constant (8.314 J K$^{-1}$ mol$^{-1}$), and T is the temperature in kelvin (25°C = 298 K).
π = 0.004 M x 8.314 J K$^{-1}$ mol$^{-1}$ x 298 K
π = 9.94 J L$^{-1}$
Finally, we need to convert the units of the osmotic pressure to atmospheres (atm), which is commonly used in osmotic pressure measurements. We can do this by dividing by the conversion factor of 101.325 J L$^{-1}$ atm$^{-1}$:
π = 9.94 J L$^{-1}$ / 101.325 J L$^{-1}$ atm$^{-1}$
π = 0.098 atm
Therefore, the osmotic pressure of the solution is 0.098 atm.
First, we need to calculate the molarity of the solution:
Molarity = moles of solute / volume of solution in liters
Since we have 1.00 g of solute and its molar mass is 250 g mol$^{-1}$, we can calculate the number of moles:
moles of solute = 1.00 g / 250 g mol$^{-1}$ = 0.004 mol
The volume of the solution is given as 1000 mL, which is equal to 1.00 L. Therefore, the molarity of the solution is:
Molarity = 0.004 mol / 1.00 L = 0.004 M
Next, we can use the van't Hoff equation to calculate the osmotic pressure:
π = MRT
where π is the osmotic pressure, M is the molarity of the solution, R is the gas constant (8.314 J K$^{-1}$ mol$^{-1}$), and T is the temperature in kelvin (25°C = 298 K).
π = 0.004 M x 8.314 J K$^{-1}$ mol$^{-1}$ x 298 K
π = 9.94 J L$^{-1}$
Finally, we need to convert the units of the osmotic pressure to atmospheres (atm), which is commonly used in osmotic pressure measurements. We can do this by dividing by the conversion factor of 101.325 J L$^{-1}$ atm$^{-1}$:
π = 9.94 J L$^{-1}$ / 101.325 J L$^{-1}$ atm$^{-1}$
π = 0.098 atm
Therefore, the osmotic pressure of the solution is 0.098 atm.
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1.00 g of a non-electrolyte solute (molar mass 250 g mol–1) was dissolved in 51.2 g of benzene. If the freezing point depression constant, Kf of benzene is 5.12 K kg mol–1, the freezing point of benzene will be lowered by [2006]a)0.3 Kb)0.5 Kc)0.4 Kd)0.2Correct answer is option 'C'. Can you explain this answer?
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1.00 g of a non-electrolyte solute (molar mass 250 g mol–1) was dissolved in 51.2 g of benzene. If the freezing point depression constant, Kf of benzene is 5.12 K kg mol–1, the freezing point of benzene will be lowered by [2006]a)0.3 Kb)0.5 Kc)0.4 Kd)0.2Correct answer is option 'C'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about 1.00 g of a non-electrolyte solute (molar mass 250 g mol–1) was dissolved in 51.2 g of benzene. If the freezing point depression constant, Kf of benzene is 5.12 K kg mol–1, the freezing point of benzene will be lowered by [2006]a)0.3 Kb)0.5 Kc)0.4 Kd)0.2Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1.00 g of a non-electrolyte solute (molar mass 250 g mol–1) was dissolved in 51.2 g of benzene. If the freezing point depression constant, Kf of benzene is 5.12 K kg mol–1, the freezing point of benzene will be lowered by [2006]a)0.3 Kb)0.5 Kc)0.4 Kd)0.2Correct answer is option 'C'. Can you explain this answer?.
1.00 g of a non-electrolyte solute (molar mass 250 g mol–1) was dissolved in 51.2 g of benzene. If the freezing point depression constant, Kf of benzene is 5.12 K kg mol–1, the freezing point of benzene will be lowered by [2006]a)0.3 Kb)0.5 Kc)0.4 Kd)0.2Correct answer is option 'C'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about 1.00 g of a non-electrolyte solute (molar mass 250 g mol–1) was dissolved in 51.2 g of benzene. If the freezing point depression constant, Kf of benzene is 5.12 K kg mol–1, the freezing point of benzene will be lowered by [2006]a)0.3 Kb)0.5 Kc)0.4 Kd)0.2Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1.00 g of a non-electrolyte solute (molar mass 250 g mol–1) was dissolved in 51.2 g of benzene. If the freezing point depression constant, Kf of benzene is 5.12 K kg mol–1, the freezing point of benzene will be lowered by [2006]a)0.3 Kb)0.5 Kc)0.4 Kd)0.2Correct answer is option 'C'. Can you explain this answer?.
Solutions for 1.00 g of a non-electrolyte solute (molar mass 250 g mol–1) was dissolved in 51.2 g of benzene. If the freezing point depression constant, Kf of benzene is 5.12 K kg mol–1, the freezing point of benzene will be lowered by [2006]a)0.3 Kb)0.5 Kc)0.4 Kd)0.2Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 12.
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