JEE Exam > JEE Questions > If a 40.65 gram sample of K2SO4 and BaSO4 is ...
Start Learning for Free
If a 40.65 gram sample of K2SO4 and BaSO4 is dissolved in 900 g of pure water to form a solution A at 57degrees,it's vapour pressure is found to be 39.6 torr while vapour pressure of pure water at 57degrees is 40 torr.density of A is 1.24 g/ml.in a different experiment when small amount of pure BaSO4 is mixed with water at 57 degrees it gives osmotic rise of 4.05*10^-5 atm.boiling point of solution A is?
Verified Answer
If a 40.65 gram sample of K2SO4 and BaSO4 is dissolved in 900 g of pur...
This question is part of UPSC exam. View all JEE courses
Most Upvoted Answer
If a 40.65 gram sample of K2SO4 and BaSO4 is dissolved in 900 g of pur...
Given Information:
- Mass of sample: 40.65 grams
- Mass of water: 900 grams
- Temperature: 57 degrees Celsius
- Vapour pressure of solution A: 39.6 torr
- Vapour pressure of pure water: 40 torr
- Density of solution A: 1.24 g/ml
- Osmotic rise for BaSO4: 4.05*10^-5 atm
Calculating Moles:
To determine the boiling point of solution A, we need to calculate the molality of the solute (K2SO4 and BaSO4) in the solution.
1. Calculate the moles of K2SO4 and BaSO4 in the 40.65 gram sample using their molar masses:
- Molar mass of K2SO4 = 174.26 g/mol
- Molar mass of BaSO4 = 233.38 g/mol
Moles of K2SO4 = (40.65 g) / (174.26 g/mol)
Moles of BaSO4 = (40.65 g) / (233.38 g/mol)
Calculating Molality:
2. Calculate the molality (moles of solute per kg of solvent) of the solution.
- Molality (m) = (moles of solute) / (mass of solvent in kg)
Mass of solvent (water) = 900 g = 0.9 kg
Molality of K2SO4 = (moles of K2SO4) / (0.9 kg)
Molality of BaSO4 = (moles of BaSO4) / (0.9 kg)
Determining the Boiling Point:
3. Use the elevation of boiling point equation to find the boiling point elevation (∆Tb):
- ∆Tb = Kb * m
Kb is the molal boiling point elevation constant for water, which is equal to 0.512 °C/m.
∆Tb for K2SO4 = Kb * (molality of K2SO4)
∆Tb for BaSO4 = Kb * (molality of BaSO4)
4. Subtract the elevation in boiling point (∆Tb) from the normal boiling point of water (100 °C) to find the boiling point of solution A:
Boiling point of solution A = 100 °C + ∆Tb
Calculating Osmotic Pressure:
5. Use the osmotic pressure equation to calculate the osmotic pressure (π):
- π = i * M * R * T
i is the van't Hoff factor, which is equal to the number of particles the solute dissociates into. For K2SO4, i = 3 (2 K+ ions and 1 SO4^2- ion). For BaSO4, i = 1 since it does not dissociate.
M is the molarity (moles of solute per liter of solution), which can be calculated using the moles of solute and the volume of solution.
R is the ideal gas constant, which is equal to 0.0821 L * atm / (mol * K).
T is
- Mass of sample: 40.65 grams
- Mass of water: 900 grams
- Temperature: 57 degrees Celsius
- Vapour pressure of solution A: 39.6 torr
- Vapour pressure of pure water: 40 torr
- Density of solution A: 1.24 g/ml
- Osmotic rise for BaSO4: 4.05*10^-5 atm
Calculating Moles:
To determine the boiling point of solution A, we need to calculate the molality of the solute (K2SO4 and BaSO4) in the solution.
1. Calculate the moles of K2SO4 and BaSO4 in the 40.65 gram sample using their molar masses:
- Molar mass of K2SO4 = 174.26 g/mol
- Molar mass of BaSO4 = 233.38 g/mol
Moles of K2SO4 = (40.65 g) / (174.26 g/mol)
Moles of BaSO4 = (40.65 g) / (233.38 g/mol)
Calculating Molality:
2. Calculate the molality (moles of solute per kg of solvent) of the solution.
- Molality (m) = (moles of solute) / (mass of solvent in kg)
Mass of solvent (water) = 900 g = 0.9 kg
Molality of K2SO4 = (moles of K2SO4) / (0.9 kg)
Molality of BaSO4 = (moles of BaSO4) / (0.9 kg)
Determining the Boiling Point:
3. Use the elevation of boiling point equation to find the boiling point elevation (∆Tb):
- ∆Tb = Kb * m
Kb is the molal boiling point elevation constant for water, which is equal to 0.512 °C/m.
∆Tb for K2SO4 = Kb * (molality of K2SO4)
∆Tb for BaSO4 = Kb * (molality of BaSO4)
4. Subtract the elevation in boiling point (∆Tb) from the normal boiling point of water (100 °C) to find the boiling point of solution A:
Boiling point of solution A = 100 °C + ∆Tb
Calculating Osmotic Pressure:
5. Use the osmotic pressure equation to calculate the osmotic pressure (π):
- π = i * M * R * T
i is the van't Hoff factor, which is equal to the number of particles the solute dissociates into. For K2SO4, i = 3 (2 K+ ions and 1 SO4^2- ion). For BaSO4, i = 1 since it does not dissociate.
M is the molarity (moles of solute per liter of solution), which can be calculated using the moles of solute and the volume of solution.
R is the ideal gas constant, which is equal to 0.0821 L * atm / (mol * K).
T is
Attention JEE Students!
To make sure you are not studying endlessly, EduRev has designed JEE study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in JEE.
|
|
Explore Courses for JEE exam
|
|
Similar JEE Doubts
Top Courses for JEEView all
If a 40.65 gram sample of K2SO4 and BaSO4 is dissolved in 900 g of pure water to form a solution A at 57degrees,it's vapour pressure is found to be 39.6 torr while vapour pressure of pure water at 57degrees is 40 torr.density of A is 1.24 g/ml.in a different experiment when small amount of pure BaSO4 is mixed with water at 57 degrees it gives osmotic rise of 4.05*10^-5 atm.boiling point of solution A is?
Question Description
If a 40.65 gram sample of K2SO4 and BaSO4 is dissolved in 900 g of pure water to form a solution A at 57degrees,it's vapour pressure is found to be 39.6 torr while vapour pressure of pure water at 57degrees is 40 torr.density of A is 1.24 g/ml.in a different experiment when small amount of pure BaSO4 is mixed with water at 57 degrees it gives osmotic rise of 4.05*10^-5 atm.boiling point of solution A is? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about If a 40.65 gram sample of K2SO4 and BaSO4 is dissolved in 900 g of pure water to form a solution A at 57degrees,it's vapour pressure is found to be 39.6 torr while vapour pressure of pure water at 57degrees is 40 torr.density of A is 1.24 g/ml.in a different experiment when small amount of pure BaSO4 is mixed with water at 57 degrees it gives osmotic rise of 4.05*10^-5 atm.boiling point of solution A is? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If a 40.65 gram sample of K2SO4 and BaSO4 is dissolved in 900 g of pure water to form a solution A at 57degrees,it's vapour pressure is found to be 39.6 torr while vapour pressure of pure water at 57degrees is 40 torr.density of A is 1.24 g/ml.in a different experiment when small amount of pure BaSO4 is mixed with water at 57 degrees it gives osmotic rise of 4.05*10^-5 atm.boiling point of solution A is?.
If a 40.65 gram sample of K2SO4 and BaSO4 is dissolved in 900 g of pure water to form a solution A at 57degrees,it's vapour pressure is found to be 39.6 torr while vapour pressure of pure water at 57degrees is 40 torr.density of A is 1.24 g/ml.in a different experiment when small amount of pure BaSO4 is mixed with water at 57 degrees it gives osmotic rise of 4.05*10^-5 atm.boiling point of solution A is? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about If a 40.65 gram sample of K2SO4 and BaSO4 is dissolved in 900 g of pure water to form a solution A at 57degrees,it's vapour pressure is found to be 39.6 torr while vapour pressure of pure water at 57degrees is 40 torr.density of A is 1.24 g/ml.in a different experiment when small amount of pure BaSO4 is mixed with water at 57 degrees it gives osmotic rise of 4.05*10^-5 atm.boiling point of solution A is? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If a 40.65 gram sample of K2SO4 and BaSO4 is dissolved in 900 g of pure water to form a solution A at 57degrees,it's vapour pressure is found to be 39.6 torr while vapour pressure of pure water at 57degrees is 40 torr.density of A is 1.24 g/ml.in a different experiment when small amount of pure BaSO4 is mixed with water at 57 degrees it gives osmotic rise of 4.05*10^-5 atm.boiling point of solution A is?.
Solutions for If a 40.65 gram sample of K2SO4 and BaSO4 is dissolved in 900 g of pure water to form a solution A at 57degrees,it's vapour pressure is found to be 39.6 torr while vapour pressure of pure water at 57degrees is 40 torr.density of A is 1.24 g/ml.in a different experiment when small amount of pure BaSO4 is mixed with water at 57 degrees it gives osmotic rise of 4.05*10^-5 atm.boiling point of solution A is? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of If a 40.65 gram sample of K2SO4 and BaSO4 is dissolved in 900 g of pure water to form a solution A at 57degrees,it's vapour pressure is found to be 39.6 torr while vapour pressure of pure water at 57degrees is 40 torr.density of A is 1.24 g/ml.in a different experiment when small amount of pure BaSO4 is mixed with water at 57 degrees it gives osmotic rise of 4.05*10^-5 atm.boiling point of solution A is? defined & explained in the simplest way possible. Besides giving the explanation of
If a 40.65 gram sample of K2SO4 and BaSO4 is dissolved in 900 g of pure water to form a solution A at 57degrees,it's vapour pressure is found to be 39.6 torr while vapour pressure of pure water at 57degrees is 40 torr.density of A is 1.24 g/ml.in a different experiment when small amount of pure BaSO4 is mixed with water at 57 degrees it gives osmotic rise of 4.05*10^-5 atm.boiling point of solution A is?, a detailed solution for If a 40.65 gram sample of K2SO4 and BaSO4 is dissolved in 900 g of pure water to form a solution A at 57degrees,it's vapour pressure is found to be 39.6 torr while vapour pressure of pure water at 57degrees is 40 torr.density of A is 1.24 g/ml.in a different experiment when small amount of pure BaSO4 is mixed with water at 57 degrees it gives osmotic rise of 4.05*10^-5 atm.boiling point of solution A is? has been provided alongside types of If a 40.65 gram sample of K2SO4 and BaSO4 is dissolved in 900 g of pure water to form a solution A at 57degrees,it's vapour pressure is found to be 39.6 torr while vapour pressure of pure water at 57degrees is 40 torr.density of A is 1.24 g/ml.in a different experiment when small amount of pure BaSO4 is mixed with water at 57 degrees it gives osmotic rise of 4.05*10^-5 atm.boiling point of solution A is? theory, EduRev gives you an
ample number of questions to practice If a 40.65 gram sample of K2SO4 and BaSO4 is dissolved in 900 g of pure water to form a solution A at 57degrees,it's vapour pressure is found to be 39.6 torr while vapour pressure of pure water at 57degrees is 40 torr.density of A is 1.24 g/ml.in a different experiment when small amount of pure BaSO4 is mixed with water at 57 degrees it gives osmotic rise of 4.05*10^-5 atm.boiling point of solution A is? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up
within 7 days!
within 7 days!
Takes less than 10 seconds to signup