Class 11 Exam > Class 11 Questions > An aqueous solution of Na2SO4 , contains 30% ...
Start Learning for Free
An aqueous solution of Na2SO4 , contains 30% Na2SO4 by mass. Density of solution is 1.6 g/ml . Calculate the no. of electrons in 62.5 ml of this solution?
Most Upvoted Answer
An aqueous solution of Na2SO4 , contains 30% Na2SO4 by mass. Density o...
Community Answer
An aqueous solution of Na2SO4 , contains 30% Na2SO4 by mass. Density o...
Calculating the Number of Electrons in the Solution
Given Data:
- Mass percentage of Na2SO4 in solution: 30%
- Density of the solution: 1.6 g/ml
- Volume of the solution: 62.5 ml
Step 1: Calculate the Mass of Na2SO4 in 62.5 ml of Solution
- Mass of 62.5 ml solution = Volume x Density = 62.5 ml x 1.6 g/ml = 100 g
- Mass of Na2SO4 in solution = 30% of 100 g = 30 g
Step 2: Determine the Number of Moles of Na2SO4
- Molar mass of Na2SO4 = 2(Na) + 1(S) + 4(O) = 2(23) + 32 + 4(16) = 142 g/mol
- Number of moles of Na2SO4 = Mass / Molar mass = 30 g / 142 g/mol ≈ 0.211 moles
Step 3: Find the Number of Electrons in the Solution
- In 1 molecule of Na2SO4, there are 10 electrons (2 from Na, 6 from S, and 2 from O)
- Number of electrons in 0.211 moles of Na2SO4 = 0.211 moles x 6.022 x 10^23 molecules/mol x 10 electrons/molecule
Step 4: Calculate the Final Result
- Number of electrons in 62.5 ml of solution ≈ 0.211 x 6.022 x 10^23 x 10 ≈ 1.27 x 10^23 electrons
Therefore, there are approximately 1.27 x 10^23 electrons in 62.5 ml of the given Na2SO4 solution.
Given Data:
- Mass percentage of Na2SO4 in solution: 30%
- Density of the solution: 1.6 g/ml
- Volume of the solution: 62.5 ml
Step 1: Calculate the Mass of Na2SO4 in 62.5 ml of Solution
- Mass of 62.5 ml solution = Volume x Density = 62.5 ml x 1.6 g/ml = 100 g
- Mass of Na2SO4 in solution = 30% of 100 g = 30 g
Step 2: Determine the Number of Moles of Na2SO4
- Molar mass of Na2SO4 = 2(Na) + 1(S) + 4(O) = 2(23) + 32 + 4(16) = 142 g/mol
- Number of moles of Na2SO4 = Mass / Molar mass = 30 g / 142 g/mol ≈ 0.211 moles
Step 3: Find the Number of Electrons in the Solution
- In 1 molecule of Na2SO4, there are 10 electrons (2 from Na, 6 from S, and 2 from O)
- Number of electrons in 0.211 moles of Na2SO4 = 0.211 moles x 6.022 x 10^23 molecules/mol x 10 electrons/molecule
Step 4: Calculate the Final Result
- Number of electrons in 62.5 ml of solution ≈ 0.211 x 6.022 x 10^23 x 10 ≈ 1.27 x 10^23 electrons
Therefore, there are approximately 1.27 x 10^23 electrons in 62.5 ml of the given Na2SO4 solution.
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
An aqueous solution of Na2SO4 , contains 30% Na2SO4 by mass. Density of solution is 1.6 g/ml . Calculate the no. of electrons in 62.5 ml of this solution?
Question Description
An aqueous solution of Na2SO4 , contains 30% Na2SO4 by mass. Density of solution is 1.6 g/ml . Calculate the no. of electrons in 62.5 ml of this solution? 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 An aqueous solution of Na2SO4 , contains 30% Na2SO4 by mass. Density of solution is 1.6 g/ml . Calculate the no. of electrons in 62.5 ml of this solution? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An aqueous solution of Na2SO4 , contains 30% Na2SO4 by mass. Density of solution is 1.6 g/ml . Calculate the no. of electrons in 62.5 ml of this solution?.
An aqueous solution of Na2SO4 , contains 30% Na2SO4 by mass. Density of solution is 1.6 g/ml . Calculate the no. of electrons in 62.5 ml of this solution? 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 An aqueous solution of Na2SO4 , contains 30% Na2SO4 by mass. Density of solution is 1.6 g/ml . Calculate the no. of electrons in 62.5 ml of this solution? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An aqueous solution of Na2SO4 , contains 30% Na2SO4 by mass. Density of solution is 1.6 g/ml . Calculate the no. of electrons in 62.5 ml of this solution?.
Solutions for An aqueous solution of Na2SO4 , contains 30% Na2SO4 by mass. Density of solution is 1.6 g/ml . Calculate the no. of electrons in 62.5 ml of this solution? 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 An aqueous solution of Na2SO4 , contains 30% Na2SO4 by mass. Density of solution is 1.6 g/ml . Calculate the no. of electrons in 62.5 ml of this solution? defined & explained in the simplest way possible. Besides giving the explanation of
An aqueous solution of Na2SO4 , contains 30% Na2SO4 by mass. Density of solution is 1.6 g/ml . Calculate the no. of electrons in 62.5 ml of this solution?, a detailed solution for An aqueous solution of Na2SO4 , contains 30% Na2SO4 by mass. Density of solution is 1.6 g/ml . Calculate the no. of electrons in 62.5 ml of this solution? has been provided alongside types of An aqueous solution of Na2SO4 , contains 30% Na2SO4 by mass. Density of solution is 1.6 g/ml . Calculate the no. of electrons in 62.5 ml of this solution? theory, EduRev gives you an
ample number of questions to practice An aqueous solution of Na2SO4 , contains 30% Na2SO4 by mass. Density of solution is 1.6 g/ml . Calculate the no. of electrons in 62.5 ml of this solution? tests, examples and also practice Class 11 tests.
|
|
Explore Courses for Class 11 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
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup