Class 11 Exam > Class 11 Questions > On evaporation of 1L of urea solutions of den...
Start Learning for Free
On evaporation of 1L of urea solutions of density 1.2g/ml, 120 g residue is formed. Thus, molality of solution is A. 2000 molal B. 1.825 molal C. 1.000 molal D. 1.200 molal?
Most Upvoted Answer
On evaporation of 1L of urea solutions of density 1.2g/ml, 120 g resid...
Community Answer
On evaporation of 1L of urea solutions of density 1.2g/ml, 120 g resid...
Given data:
- Volume of urea solution = 1 L
- Density of urea solution = 1.2 g/mL
- Mass of residue formed = 120 g
Calculating the mass of the solution:
Mass = Density × Volume
Mass of solution = 1.2 g/mL × 1000 mL = 1200 g
Calculating the mass of urea:
Mass of urea = Mass of solution - Mass of residue
Mass of urea = 1200 g - 120 g = 1080 g
Calculating the number of moles of urea:
Molar mass of urea (CH4N2O) = 12.01 g/mol + 4(1.01 g/mol) + 2(14.01 g/mol) + 16.00 g/mol = 60.06 g/mol
Number of moles of urea = Mass of urea / Molar mass of urea
Number of moles of urea = 1080 g / 60.06 g/mol = 17.97 mol
Calculating the molality of the solution:
Molality (m) = Number of moles of solute / Mass of solvent in kg
Since the solvent is water, its mass is equal to its volume.
Mass of solvent = Volume of urea solution × Density of urea solution
Mass of solvent = 1 L × 1.2 g/mL = 1200 g = 1.2 kg
Molality (m) = 17.97 mol / 1.2 kg = 14.975 mol/kg ≈ 1.825 molal
Answer:
The molality of the urea solution is approximately 1.825 molal. Therefore, the correct option is B. 1.825 molal.
- Volume of urea solution = 1 L
- Density of urea solution = 1.2 g/mL
- Mass of residue formed = 120 g
Calculating the mass of the solution:
Mass = Density × Volume
Mass of solution = 1.2 g/mL × 1000 mL = 1200 g
Calculating the mass of urea:
Mass of urea = Mass of solution - Mass of residue
Mass of urea = 1200 g - 120 g = 1080 g
Calculating the number of moles of urea:
Molar mass of urea (CH4N2O) = 12.01 g/mol + 4(1.01 g/mol) + 2(14.01 g/mol) + 16.00 g/mol = 60.06 g/mol
Number of moles of urea = Mass of urea / Molar mass of urea
Number of moles of urea = 1080 g / 60.06 g/mol = 17.97 mol
Calculating the molality of the solution:
Molality (m) = Number of moles of solute / Mass of solvent in kg
Since the solvent is water, its mass is equal to its volume.
Mass of solvent = Volume of urea solution × Density of urea solution
Mass of solvent = 1 L × 1.2 g/mL = 1200 g = 1.2 kg
Molality (m) = 17.97 mol / 1.2 kg = 14.975 mol/kg ≈ 1.825 molal
Answer:
The molality of the urea solution is approximately 1.825 molal. Therefore, the correct option is B. 1.825 molal.
Attention Class 11 Students!
To make sure you are not studying endlessly, EduRev has designed Class 11 study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Class 11.
|
Explore Courses for Class 11 exam
|
|
Similar Class 11 Doubts
Top Courses for Class 11View all
On evaporation of 1L of urea solutions of density 1.2g/ml, 120 g residue is formed. Thus, molality of solution is A. 2000 molal B. 1.825 molal C. 1.000 molal D. 1.200 molal?
Question Description
On evaporation of 1L of urea solutions of density 1.2g/ml, 120 g residue is formed. Thus, molality of solution is A. 2000 molal B. 1.825 molal C. 1.000 molal D. 1.200 molal? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about On evaporation of 1L of urea solutions of density 1.2g/ml, 120 g residue is formed. Thus, molality of solution is A. 2000 molal B. 1.825 molal C. 1.000 molal D. 1.200 molal? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for On evaporation of 1L of urea solutions of density 1.2g/ml, 120 g residue is formed. Thus, molality of solution is A. 2000 molal B. 1.825 molal C. 1.000 molal D. 1.200 molal?.
On evaporation of 1L of urea solutions of density 1.2g/ml, 120 g residue is formed. Thus, molality of solution is A. 2000 molal B. 1.825 molal C. 1.000 molal D. 1.200 molal? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about On evaporation of 1L of urea solutions of density 1.2g/ml, 120 g residue is formed. Thus, molality of solution is A. 2000 molal B. 1.825 molal C. 1.000 molal D. 1.200 molal? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for On evaporation of 1L of urea solutions of density 1.2g/ml, 120 g residue is formed. Thus, molality of solution is A. 2000 molal B. 1.825 molal C. 1.000 molal D. 1.200 molal?.
Solutions for On evaporation of 1L of urea solutions of density 1.2g/ml, 120 g residue is formed. Thus, molality of solution is A. 2000 molal B. 1.825 molal C. 1.000 molal D. 1.200 molal? in English & in Hindi are available as part of our courses for Class 11.
Download more important topics, notes, lectures and mock test series for Class 11 Exam by signing up for free.
Here you can find the meaning of On evaporation of 1L of urea solutions of density 1.2g/ml, 120 g residue is formed. Thus, molality of solution is A. 2000 molal B. 1.825 molal C. 1.000 molal D. 1.200 molal? defined & explained in the simplest way possible. Besides giving the explanation of
On evaporation of 1L of urea solutions of density 1.2g/ml, 120 g residue is formed. Thus, molality of solution is A. 2000 molal B. 1.825 molal C. 1.000 molal D. 1.200 molal?, a detailed solution for On evaporation of 1L of urea solutions of density 1.2g/ml, 120 g residue is formed. Thus, molality of solution is A. 2000 molal B. 1.825 molal C. 1.000 molal D. 1.200 molal? has been provided alongside types of On evaporation of 1L of urea solutions of density 1.2g/ml, 120 g residue is formed. Thus, molality of solution is A. 2000 molal B. 1.825 molal C. 1.000 molal D. 1.200 molal? theory, EduRev gives you an
ample number of questions to practice On evaporation of 1L of urea solutions of density 1.2g/ml, 120 g residue is formed. Thus, molality of solution is A. 2000 molal B. 1.825 molal C. 1.000 molal D. 1.200 molal? tests, examples and also practice Class 11 tests.
|
|
Explore Courses for Class 11 exam
|
|
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!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup