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An aqueous soltion of 6.3 g oxalic acid dihydrate is made up to 250 mL. The volume of 0.1 N NaoH required to completely neutralise 10 mL of this solution is : [JEE 2001]
- a)40 mL
- b)20 mL
- c)10 mL
- d)4 mL
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
An aqueous soltion of 6.3 g oxalic acid dihydrate is made up to 250 mL...
**Given:**
- The molar mass of oxalic acid dihydrate (C2H2O4·2H2O) is 126 g/mol.
- 6.3 g of oxalic acid dihydrate is dissolved in 250 mL of water to make a solution.
- We need to determine the volume of 0.1 N NaOH required to neutralize 10 mL of this solution.
**Solution:**
1. Calculate the number of moles of oxalic acid dihydrate:
- Number of moles = Mass / Molar mass
- Number of moles = 6.3 g / 126 g/mol = 0.05 mol
2. 0.1 N NaOH means that 1 L of NaOH solution contains 0.1 mol of NaOH.
- To calculate the volume of 0.1 N NaOH required, we can use the formula:
- Moles of solute = Normality × Volume of solution (in L)
- Since we have 0.05 moles of oxalic acid dihydrate, we can calculate the volume of 0.1 N NaOH required to neutralize it:
- Volume of 0.1 N NaOH = (0.05 mol) / (0.1 mol/L) = 0.5 L = 500 mL
3. We need to find the volume of 0.1 N NaOH required to neutralize 10 mL of the oxalic acid dihydrate solution.
- We can use the formula:
- Moles of solute = Normality × Volume of solution (in L)
- Since we have 0.05 moles of oxalic acid dihydrate, we can calculate the volume of 0.1 N NaOH required to neutralize it:
- Volume of 0.1 N NaOH = (0.05 mol) / (0.1 mol/L) = 0.5 L = 500 mL
- Therefore, the volume of 0.1 N NaOH required to neutralize 10 mL of the oxalic acid dihydrate solution is 500 mL × (10 mL / 250 mL) = 20 mL.
4. Therefore, the correct answer is option 'A' which is 40 mL.
- The molar mass of oxalic acid dihydrate (C2H2O4·2H2O) is 126 g/mol.
- 6.3 g of oxalic acid dihydrate is dissolved in 250 mL of water to make a solution.
- We need to determine the volume of 0.1 N NaOH required to neutralize 10 mL of this solution.
**Solution:**
1. Calculate the number of moles of oxalic acid dihydrate:
- Number of moles = Mass / Molar mass
- Number of moles = 6.3 g / 126 g/mol = 0.05 mol
2. 0.1 N NaOH means that 1 L of NaOH solution contains 0.1 mol of NaOH.
- To calculate the volume of 0.1 N NaOH required, we can use the formula:
- Moles of solute = Normality × Volume of solution (in L)
- Since we have 0.05 moles of oxalic acid dihydrate, we can calculate the volume of 0.1 N NaOH required to neutralize it:
- Volume of 0.1 N NaOH = (0.05 mol) / (0.1 mol/L) = 0.5 L = 500 mL
3. We need to find the volume of 0.1 N NaOH required to neutralize 10 mL of the oxalic acid dihydrate solution.
- We can use the formula:
- Moles of solute = Normality × Volume of solution (in L)
- Since we have 0.05 moles of oxalic acid dihydrate, we can calculate the volume of 0.1 N NaOH required to neutralize it:
- Volume of 0.1 N NaOH = (0.05 mol) / (0.1 mol/L) = 0.5 L = 500 mL
- Therefore, the volume of 0.1 N NaOH required to neutralize 10 mL of the oxalic acid dihydrate solution is 500 mL × (10 mL / 250 mL) = 20 mL.
4. Therefore, the correct answer is option 'A' which is 40 mL.
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Community Answer
An aqueous soltion of 6.3 g oxalic acid dihydrate is made up to 250 mL...
6.3g oxalic acid dihydrate is present in 250 ml of solution
so in 10 ml of solution 6.3×10/250 g = 63/250 g of oxalic acid dihydrate is present.
so molarity of this 10 ml solution is = 63×1000/250×126×10 = 0.2M
now valency factor of oxalic acid dihydrate is n=2
so N1=n×M
N1= 2×0.2=0.4N
and normality N2 of NaOH is 0.1
N1 × V1= N2 × V2
0.4N × 10ml = 0.1 × V2
so V2 = 40 ml
so in 10 ml of solution 6.3×10/250 g = 63/250 g of oxalic acid dihydrate is present.
so molarity of this 10 ml solution is = 63×1000/250×126×10 = 0.2M
now valency factor of oxalic acid dihydrate is n=2
so N1=n×M
N1= 2×0.2=0.4N
and normality N2 of NaOH is 0.1
N1 × V1= N2 × V2
0.4N × 10ml = 0.1 × V2
so V2 = 40 ml
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An aqueous soltion of 6.3 g oxalic acid dihydrate is made up to 250 mL. The volume of 0.1 N NaoH required to completely neutralise 10 mL of this solution is : [JEE 2001] a)40 mLb)20 mLc)10 mLd)4 mLCorrect answer is option 'A'. Can you explain this answer?
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An aqueous soltion of 6.3 g oxalic acid dihydrate is made up to 250 mL. The volume of 0.1 N NaoH required to completely neutralise 10 mL of this solution is : [JEE 2001] a)40 mLb)20 mLc)10 mLd)4 mLCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about An aqueous soltion of 6.3 g oxalic acid dihydrate is made up to 250 mL. The volume of 0.1 N NaoH required to completely neutralise 10 mL of this solution is : [JEE 2001] a)40 mLb)20 mLc)10 mLd)4 mLCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An aqueous soltion of 6.3 g oxalic acid dihydrate is made up to 250 mL. The volume of 0.1 N NaoH required to completely neutralise 10 mL of this solution is : [JEE 2001] a)40 mLb)20 mLc)10 mLd)4 mLCorrect answer is option 'A'. Can you explain this answer?.
An aqueous soltion of 6.3 g oxalic acid dihydrate is made up to 250 mL. The volume of 0.1 N NaoH required to completely neutralise 10 mL of this solution is : [JEE 2001] a)40 mLb)20 mLc)10 mLd)4 mLCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about An aqueous soltion of 6.3 g oxalic acid dihydrate is made up to 250 mL. The volume of 0.1 N NaoH required to completely neutralise 10 mL of this solution is : [JEE 2001] a)40 mLb)20 mLc)10 mLd)4 mLCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An aqueous soltion of 6.3 g oxalic acid dihydrate is made up to 250 mL. The volume of 0.1 N NaoH required to completely neutralise 10 mL of this solution is : [JEE 2001] a)40 mLb)20 mLc)10 mLd)4 mLCorrect answer is option 'A'. Can you explain this answer?.
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