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An aquous solution of 6.3 gram oxalic acid is made up to 250 ml. The volume of 0.1 N NaOH required to completely neutralized 10 ml of the solution is.?
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An aquous solution of 6.3 gram oxalic acid is made up to 250 ml. The v...
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An aquous solution of 6.3 gram oxalic acid is made up to 250 ml. The v...
Calculation of Volume of NaOH Required
- First, calculate the molarity of the oxalic acid solution:
- Number of moles of oxalic acid = (6.3 g) / (Molar mass of oxalic acid)
- Molar mass of oxalic acid = 90.03 g/mol
- Number of moles of oxalic acid = (6.3 g) / (90.03 g/mol) = 0.07 mol
- Molarity of oxalic acid solution = (0.07 mol) / (0.25 L) = 0.28 M
- Now, calculate the volume of 0.1 N NaOH required to neutralize 10 ml of the oxalic acid solution:
- Number of moles of oxalic acid in 10 ml = (0.28 mol/L) x (0.01 L) = 0.0028 mol
- Since oxalic acid is a diprotic acid, it requires twice the amount of NaOH for neutralization
- Number of moles of NaOH required = 2 x 0.0028 mol = 0.0056 mol
- Calculate the volume of 0.1 N NaOH needed:
- Molarity of NaOH = 0.1 N = 0.1 mol/L
- Volume of NaOH required = (0.0056 mol) / (0.1 mol/L) = 0.056 L = 56 ml
Therefore, 56 ml of 0.1 N NaOH is required to completely neutralize 10 ml of the oxalic acid solution.
- First, calculate the molarity of the oxalic acid solution:
- Number of moles of oxalic acid = (6.3 g) / (Molar mass of oxalic acid)
- Molar mass of oxalic acid = 90.03 g/mol
- Number of moles of oxalic acid = (6.3 g) / (90.03 g/mol) = 0.07 mol
- Molarity of oxalic acid solution = (0.07 mol) / (0.25 L) = 0.28 M
- Now, calculate the volume of 0.1 N NaOH required to neutralize 10 ml of the oxalic acid solution:
- Number of moles of oxalic acid in 10 ml = (0.28 mol/L) x (0.01 L) = 0.0028 mol
- Since oxalic acid is a diprotic acid, it requires twice the amount of NaOH for neutralization
- Number of moles of NaOH required = 2 x 0.0028 mol = 0.0056 mol
- Calculate the volume of 0.1 N NaOH needed:
- Molarity of NaOH = 0.1 N = 0.1 mol/L
- Volume of NaOH required = (0.0056 mol) / (0.1 mol/L) = 0.056 L = 56 ml
Therefore, 56 ml of 0.1 N NaOH is required to completely neutralize 10 ml of the oxalic acid solution.
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An aquous solution of 6.3 gram oxalic acid is made up to 250 ml. The volume of 0.1 N NaOH required to completely neutralized 10 ml of the solution is.?
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An aquous solution of 6.3 gram oxalic acid is made up to 250 ml. The volume of 0.1 N NaOH required to completely neutralized 10 ml of the solution is.? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about An aquous solution of 6.3 gram oxalic acid is made up to 250 ml. The volume of 0.1 N NaOH required to completely neutralized 10 ml of the solution is.? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An aquous solution of 6.3 gram oxalic acid is made up to 250 ml. The volume of 0.1 N NaOH required to completely neutralized 10 ml of the solution is.?.
An aquous solution of 6.3 gram oxalic acid is made up to 250 ml. The volume of 0.1 N NaOH required to completely neutralized 10 ml of the solution is.? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about An aquous solution of 6.3 gram oxalic acid is made up to 250 ml. The volume of 0.1 N NaOH required to completely neutralized 10 ml of the solution is.? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An aquous solution of 6.3 gram oxalic acid is made up to 250 ml. The volume of 0.1 N NaOH required to completely neutralized 10 ml of the solution is.?.
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