IIT JAM Exam > IIT JAM Questions > A mixture of H2SO4 and H2C2O4 (oxalic acid) a...
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A mixture of H2SO4 and H2C2O4 (oxalic acid) and some inert impurity weighing 3.185 g was dissolved in water and the solution made up to 1 litre. 10 mL of this solution required 3 mL of 0.1 N NaOH for complete neutralization. In another experiment, 100 mL of the same solution in hot condition required 4 mL of 0.02 M KMnO4 solution for complete reaction. The % by weight of H2SO4 in the mixture was:
- a)40
- b)50
- c)60
- d)80
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
A mixture of H2SO4 and H2C2O4 (oxalic acid) and some inert impurity we...
To determine the percentage by weight of H2SO4 in the mixture, we can use the concept of normality and stoichiometry.
Given information:
- A mixture of H2SO4 and H2C2O4 (oxalic acid) and some inert impurity weighing 3.185 g was dissolved in water and made up to 1 litre.
- 10 mL of this solution required 3 mL of 0.1 N NaOH for complete neutralization.
- 100 mL of the same solution in hot condition required 4 mL of 0.02 M KMnO4 solution for complete reaction.
Let's solve this step by step:
Step 1: Determine the normality of H2SO4 in the given solution.
- From the neutralization reaction between H2SO4 and NaOH, we can determine the normality of H2SO4.
- According to the equation: H2SO4 + 2NaOH → Na2SO4 + 2H2O
- The mole ratio between H2SO4 and NaOH is 1:2.
- From the given information, 10 mL of the solution required 3 mL of 0.1 N NaOH.
- Using the equation, we can calculate the normality of H2SO4:
Normality of H2SO4 = Normality of NaOH × Volume of NaOH / Volume of H2SO4
Normality of H2SO4 = 0.1 N × 3 mL / 10 mL
Normality of H2SO4 = 0.03 N
Step 2: Determine the moles of H2SO4 in the given solution.
- The normality of a solution is defined as the number of gram equivalents of solute present in one liter of the solution.
- The gram equivalent weight of H2SO4 is equal to its molar mass divided by the number of acidic hydrogen atoms.
- The molar mass of H2SO4 is 98 g/mol, and it has 2 acidic hydrogen atoms.
- Therefore, the gram equivalent weight of H2SO4 is 98 g/mol / 2 = 49 g/equivalent.
- The moles of H2SO4 can be calculated using the formula:
Moles of H2SO4 = Normality of H2SO4 × Volume of H2SO4 / 1000
Moles of H2SO4 = 0.03 N × 10 mL / 1000 mL
Moles of H2SO4 = 0.0003 moles
Step 3: Determine the moles of oxalic acid (H2C2O4) in the given solution.
- Since the volume of oxalic acid is not given, we can assume it to be the remaining volume of the solution after adding the H2SO4 and inert impurity.
- Let the volume of oxalic acid be V mL.
- The total volume of the solution is 1000 mL.
- Therefore, the volume of oxalic acid is V mL = 1000 mL - 10 mL (H2SO4 volume)
- The normality of oxalic acid can be calculated using the formula:
Normality of H2C2O4 = Normality of H2SO4 × Volume of H2SO4 / Volume of H2
Given information:
- A mixture of H2SO4 and H2C2O4 (oxalic acid) and some inert impurity weighing 3.185 g was dissolved in water and made up to 1 litre.
- 10 mL of this solution required 3 mL of 0.1 N NaOH for complete neutralization.
- 100 mL of the same solution in hot condition required 4 mL of 0.02 M KMnO4 solution for complete reaction.
Let's solve this step by step:
Step 1: Determine the normality of H2SO4 in the given solution.
- From the neutralization reaction between H2SO4 and NaOH, we can determine the normality of H2SO4.
- According to the equation: H2SO4 + 2NaOH → Na2SO4 + 2H2O
- The mole ratio between H2SO4 and NaOH is 1:2.
- From the given information, 10 mL of the solution required 3 mL of 0.1 N NaOH.
- Using the equation, we can calculate the normality of H2SO4:
Normality of H2SO4 = Normality of NaOH × Volume of NaOH / Volume of H2SO4
Normality of H2SO4 = 0.1 N × 3 mL / 10 mL
Normality of H2SO4 = 0.03 N
Step 2: Determine the moles of H2SO4 in the given solution.
- The normality of a solution is defined as the number of gram equivalents of solute present in one liter of the solution.
- The gram equivalent weight of H2SO4 is equal to its molar mass divided by the number of acidic hydrogen atoms.
- The molar mass of H2SO4 is 98 g/mol, and it has 2 acidic hydrogen atoms.
- Therefore, the gram equivalent weight of H2SO4 is 98 g/mol / 2 = 49 g/equivalent.
- The moles of H2SO4 can be calculated using the formula:
Moles of H2SO4 = Normality of H2SO4 × Volume of H2SO4 / 1000
Moles of H2SO4 = 0.03 N × 10 mL / 1000 mL
Moles of H2SO4 = 0.0003 moles
Step 3: Determine the moles of oxalic acid (H2C2O4) in the given solution.
- Since the volume of oxalic acid is not given, we can assume it to be the remaining volume of the solution after adding the H2SO4 and inert impurity.
- Let the volume of oxalic acid be V mL.
- The total volume of the solution is 1000 mL.
- Therefore, the volume of oxalic acid is V mL = 1000 mL - 10 mL (H2SO4 volume)
- The normality of oxalic acid can be calculated using the formula:
Normality of H2C2O4 = Normality of H2SO4 × Volume of H2SO4 / Volume of H2
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A mixture of H2SO4 and H2C2O4 (oxalic acid) and some inert impurity weighing 3.185 g was dissolved in water and the solution made up to 1 litre. 10 mL of this solution required 3 mL of 0.1 N NaOH for complete neutralization. In another experiment, 100 mL of the same solution in hot condition required 4 mL of 0.02 M KMnO4 solution for complete reaction. The % by weight of H2SO4 in the mixture was:a)40b)50c)60d)80Correct answer is option 'A'. Can you explain this answer?
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A mixture of H2SO4 and H2C2O4 (oxalic acid) and some inert impurity weighing 3.185 g was dissolved in water and the solution made up to 1 litre. 10 mL of this solution required 3 mL of 0.1 N NaOH for complete neutralization. In another experiment, 100 mL of the same solution in hot condition required 4 mL of 0.02 M KMnO4 solution for complete reaction. The % by weight of H2SO4 in the mixture was:a)40b)50c)60d)80Correct answer is option 'A'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about A mixture of H2SO4 and H2C2O4 (oxalic acid) and some inert impurity weighing 3.185 g was dissolved in water and the solution made up to 1 litre. 10 mL of this solution required 3 mL of 0.1 N NaOH for complete neutralization. In another experiment, 100 mL of the same solution in hot condition required 4 mL of 0.02 M KMnO4 solution for complete reaction. The % by weight of H2SO4 in the mixture was:a)40b)50c)60d)80Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A mixture of H2SO4 and H2C2O4 (oxalic acid) and some inert impurity weighing 3.185 g was dissolved in water and the solution made up to 1 litre. 10 mL of this solution required 3 mL of 0.1 N NaOH for complete neutralization. In another experiment, 100 mL of the same solution in hot condition required 4 mL of 0.02 M KMnO4 solution for complete reaction. The % by weight of H2SO4 in the mixture was:a)40b)50c)60d)80Correct answer is option 'A'. Can you explain this answer?.
A mixture of H2SO4 and H2C2O4 (oxalic acid) and some inert impurity weighing 3.185 g was dissolved in water and the solution made up to 1 litre. 10 mL of this solution required 3 mL of 0.1 N NaOH for complete neutralization. In another experiment, 100 mL of the same solution in hot condition required 4 mL of 0.02 M KMnO4 solution for complete reaction. The % by weight of H2SO4 in the mixture was:a)40b)50c)60d)80Correct answer is option 'A'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about A mixture of H2SO4 and H2C2O4 (oxalic acid) and some inert impurity weighing 3.185 g was dissolved in water and the solution made up to 1 litre. 10 mL of this solution required 3 mL of 0.1 N NaOH for complete neutralization. In another experiment, 100 mL of the same solution in hot condition required 4 mL of 0.02 M KMnO4 solution for complete reaction. The % by weight of H2SO4 in the mixture was:a)40b)50c)60d)80Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A mixture of H2SO4 and H2C2O4 (oxalic acid) and some inert impurity weighing 3.185 g was dissolved in water and the solution made up to 1 litre. 10 mL of this solution required 3 mL of 0.1 N NaOH for complete neutralization. In another experiment, 100 mL of the same solution in hot condition required 4 mL of 0.02 M KMnO4 solution for complete reaction. The % by weight of H2SO4 in the mixture was:a)40b)50c)60d)80Correct answer is option 'A'. Can you explain this answer?.
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ample number of questions to practice A mixture of H2SO4 and H2C2O4 (oxalic acid) and some inert impurity weighing 3.185 g was dissolved in water and the solution made up to 1 litre. 10 mL of this solution required 3 mL of 0.1 N NaOH for complete neutralization. In another experiment, 100 mL of the same solution in hot condition required 4 mL of 0.02 M KMnO4 solution for complete reaction. The % by weight of H2SO4 in the mixture was:a)40b)50c)60d)80Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice IIT JAM tests.
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