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The percentage weight of Zn in white vitriol [ZnSO4.7H2O] is approximately equal to ( Zn = 65, S = 32, O = 16 and H = 1) [1995]
- a)33.65 %
- b)32.56 %
- c)23.65 %
- d)22.65 %
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
The percentage weight of Zn in white vitriol [ZnSO4.7H2O] is approxima...
Molecular weight of ZnSO4 .7H 2O = 65 + 32 + (4 × 16) + 7(2 × 1 + 16) = 287.
∴ percentage mass of zinc (Zn)
∴ percentage mass of zinc (Zn)
Most Upvoted Answer
The percentage weight of Zn in white vitriol [ZnSO4.7H2O] is approxima...
**Solution:**
To determine the percentage weight of Zn in white vitriol (ZnSO4.7H2O), we need to calculate the molar mass of ZnSO4.7H2O and then find the percentage weight of Zn.
**Molar Mass Calculation:**
The molar mass of a compound is the sum of the atomic masses of all the atoms in the compound.
The atomic masses of the elements involved in this compound are:
Zn = 65 g/mol
S = 32 g/mol
O = 16 g/mol
H = 1 g/mol
For ZnSO4.7H2O, we have:
1 atom of Zn
1 atom of S
4 atoms of O
14 atoms of H
Now, let's calculate the molar mass:
Molar mass of Zn = 65 g/mol
Molar mass of S = 32 g/mol
Molar mass of O = 16 g/mol
Molar mass of H = 1 g/mol
Molar mass of ZnSO4.7H2O = (65) + (32) + (4 x 16) + (7 x 2 x 1) = 161 g/mol
**Percentage Weight Calculation:**
The percentage weight of an element in a compound is calculated by dividing the molar mass of the element by the molar mass of the compound and then multiplying by 100.
Percentage weight of Zn = (Molar mass of Zn / Molar mass of ZnSO4.7H2O) x 100
Percentage weight of Zn = (65 / 161) x 100
Percentage weight of Zn ≈ 40.37%
Therefore, the correct option is not given in the provided answer choices. The correct percentage weight of Zn in white vitriol (ZnSO4.7H2O) is approximately 40.37%, not 22.65%.
To determine the percentage weight of Zn in white vitriol (ZnSO4.7H2O), we need to calculate the molar mass of ZnSO4.7H2O and then find the percentage weight of Zn.
**Molar Mass Calculation:**
The molar mass of a compound is the sum of the atomic masses of all the atoms in the compound.
The atomic masses of the elements involved in this compound are:
Zn = 65 g/mol
S = 32 g/mol
O = 16 g/mol
H = 1 g/mol
For ZnSO4.7H2O, we have:
1 atom of Zn
1 atom of S
4 atoms of O
14 atoms of H
Now, let's calculate the molar mass:
Molar mass of Zn = 65 g/mol
Molar mass of S = 32 g/mol
Molar mass of O = 16 g/mol
Molar mass of H = 1 g/mol
Molar mass of ZnSO4.7H2O = (65) + (32) + (4 x 16) + (7 x 2 x 1) = 161 g/mol
**Percentage Weight Calculation:**
The percentage weight of an element in a compound is calculated by dividing the molar mass of the element by the molar mass of the compound and then multiplying by 100.
Percentage weight of Zn = (Molar mass of Zn / Molar mass of ZnSO4.7H2O) x 100
Percentage weight of Zn = (65 / 161) x 100
Percentage weight of Zn ≈ 40.37%
Therefore, the correct option is not given in the provided answer choices. The correct percentage weight of Zn in white vitriol (ZnSO4.7H2O) is approximately 40.37%, not 22.65%.
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The percentage weight of Zn in white vitriol [ZnSO4.7H2O] is approximately equal to ( Zn = 65, S = 32, O = 16 and H = 1) [1995]a)33.65 %b)32.56 %c)23.65 %d)22.65 %Correct answer is option 'D'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about The percentage weight of Zn in white vitriol [ZnSO4.7H2O] is approximately equal to ( Zn = 65, S = 32, O = 16 and H = 1) [1995]a)33.65 %b)32.56 %c)23.65 %d)22.65 %Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The percentage weight of Zn in white vitriol [ZnSO4.7H2O] is approximately equal to ( Zn = 65, S = 32, O = 16 and H = 1) [1995]a)33.65 %b)32.56 %c)23.65 %d)22.65 %Correct answer is option 'D'. Can you explain this answer?.
The percentage weight of Zn in white vitriol [ZnSO4.7H2O] is approximately equal to ( Zn = 65, S = 32, O = 16 and H = 1) [1995]a)33.65 %b)32.56 %c)23.65 %d)22.65 %Correct answer is option 'D'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about The percentage weight of Zn in white vitriol [ZnSO4.7H2O] is approximately equal to ( Zn = 65, S = 32, O = 16 and H = 1) [1995]a)33.65 %b)32.56 %c)23.65 %d)22.65 %Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The percentage weight of Zn in white vitriol [ZnSO4.7H2O] is approximately equal to ( Zn = 65, S = 32, O = 16 and H = 1) [1995]a)33.65 %b)32.56 %c)23.65 %d)22.65 %Correct answer is option 'D'. Can you explain this answer?.
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