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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?
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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.
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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?
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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?.
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