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An aqueous solution of 6.3 g oxalic acid dihydrate is made up to 250 ml. The volume of 0.1 N NaOH required to completely neutralize 10 ml of this solution is (2001S)
- a)40 ml
- b)20 ml
- c)10 ml
- d)4 ml
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
An aqueous solution of 6.3 g oxalic acid dihydrate is made up to 250 m...
TIPS/Formulae : Equivalents of H2C2O4.2H2O = Equivalents of NaOH (At equivalence point)
Strength of H2C2O4 . 2H2O (in g/L) =
= 25.2 g/L
Strength of H2C2O4 . 2H2O (in g/L) =
= 25.2 g/LNormality of H2C2O4 . 2H2O = 
Using normality equation : N1V1 = N2V2
(H2C2O4.2H2O) (NaOH)
(H2C2O4.2H2O) (NaOH)
0.4 × 10 = 0.1 × V2 or V2 = 
= 40 ml.
= 40 ml.Most Upvoted Answer
An aqueous solution of 6.3 g oxalic acid dihydrate is made up to 250 m...
Solution:
Given:
Mass of oxalic acid dihydrate = 6.3 g
Volume of solution = 250 ml
Volume of solution taken for neutralization = 10 ml
Normality of NaOH = 0.1 N
Step 1: Calculate the molarity of the oxalic acid dihydrate solution.
Molarity (M) = (moles of solute) / (volume of solution in liters)
To calculate the moles of solute, we need to convert the mass of oxalic acid dihydrate to moles.
The molar mass of oxalic acid dihydrate (C2H2O4·2H2O) = 126.07 g/mol
Moles of oxalic acid dihydrate = (mass of oxalic acid dihydrate) / (molar mass of oxalic acid dihydrate)
Moles of oxalic acid dihydrate = 6.3 g / 126.07 g/mol
Moles of oxalic acid dihydrate = 0.05 mol
Molarity (M) = 0.05 mol / 0.25 L
Molarity (M) = 0.2 M
Step 2: Calculate the volume of 0.1 N NaOH required to neutralize the solution.
The reaction between oxalic acid and sodium hydroxide can be represented as:
C2H2O4 + 2NaOH → Na2C2O4 + 2H2O
From the balanced equation, we can see that one mole of oxalic acid reacts with two moles of NaOH. Therefore, the mole ratio of oxalic acid to NaOH is 1:2.
Moles of NaOH required to neutralize the solution = (moles of oxalic acid dihydrate) x (mole ratio)
Moles of NaOH required to neutralize the solution = 0.05 mol x 2
Moles of NaOH required to neutralize the solution = 0.1 mol
Now, we can calculate the volume of 0.1 N NaOH required using the formula:
Volume of NaOH (in liters) = (moles of NaOH) / (molarity of NaOH)
Volume of NaOH (in liters) = 0.1 mol / 0.1 N
Volume of NaOH (in liters) = 1 L
However, we need to find the volume in milliliters (ml). Therefore, we can convert liters to milliliters by multiplying by 1000:
Volume of NaOH (in ml) = 1 L x 1000 ml/L
Volume of NaOH (in ml) = 1000 ml
Step 3: Calculate the volume of 0.1 N NaOH required to neutralize 10 ml of the oxalic acid dihydrate solution.
Since the volume of solution taken for neutralization is 10 ml, we can use the proportion method to calculate the volume of NaOH required.
Volume of NaOH / Volume of solution = Volume of NaOH required / Volume of solution taken for neutralization
Volume of NaOH = (Volume of NaOH required / Volume of solution taken for neutralization) x Volume of solution
Volume of NaOH = (1000 ml / 250 ml) x 10 ml
Volume of NaOH = 40 ml
Therefore, the volume of 0.
Given:
Mass of oxalic acid dihydrate = 6.3 g
Volume of solution = 250 ml
Volume of solution taken for neutralization = 10 ml
Normality of NaOH = 0.1 N
Step 1: Calculate the molarity of the oxalic acid dihydrate solution.
Molarity (M) = (moles of solute) / (volume of solution in liters)
To calculate the moles of solute, we need to convert the mass of oxalic acid dihydrate to moles.
The molar mass of oxalic acid dihydrate (C2H2O4·2H2O) = 126.07 g/mol
Moles of oxalic acid dihydrate = (mass of oxalic acid dihydrate) / (molar mass of oxalic acid dihydrate)
Moles of oxalic acid dihydrate = 6.3 g / 126.07 g/mol
Moles of oxalic acid dihydrate = 0.05 mol
Molarity (M) = 0.05 mol / 0.25 L
Molarity (M) = 0.2 M
Step 2: Calculate the volume of 0.1 N NaOH required to neutralize the solution.
The reaction between oxalic acid and sodium hydroxide can be represented as:
C2H2O4 + 2NaOH → Na2C2O4 + 2H2O
From the balanced equation, we can see that one mole of oxalic acid reacts with two moles of NaOH. Therefore, the mole ratio of oxalic acid to NaOH is 1:2.
Moles of NaOH required to neutralize the solution = (moles of oxalic acid dihydrate) x (mole ratio)
Moles of NaOH required to neutralize the solution = 0.05 mol x 2
Moles of NaOH required to neutralize the solution = 0.1 mol
Now, we can calculate the volume of 0.1 N NaOH required using the formula:
Volume of NaOH (in liters) = (moles of NaOH) / (molarity of NaOH)
Volume of NaOH (in liters) = 0.1 mol / 0.1 N
Volume of NaOH (in liters) = 1 L
However, we need to find the volume in milliliters (ml). Therefore, we can convert liters to milliliters by multiplying by 1000:
Volume of NaOH (in ml) = 1 L x 1000 ml/L
Volume of NaOH (in ml) = 1000 ml
Step 3: Calculate the volume of 0.1 N NaOH required to neutralize 10 ml of the oxalic acid dihydrate solution.
Since the volume of solution taken for neutralization is 10 ml, we can use the proportion method to calculate the volume of NaOH required.
Volume of NaOH / Volume of solution = Volume of NaOH required / Volume of solution taken for neutralization
Volume of NaOH = (Volume of NaOH required / Volume of solution taken for neutralization) x Volume of solution
Volume of NaOH = (1000 ml / 250 ml) x 10 ml
Volume of NaOH = 40 ml
Therefore, the volume of 0.
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An aqueous solution of 6.3 g oxalic acid dihydrate is made up to 250 ml. The volume of 0.1 N NaOH required to completely neutralize 10 ml of this solution is (2001S)a)40 mlb)20 mlc)10 mld)4 mlCorrect answer is option 'A'. Can you explain this answer?
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An aqueous solution of 6.3 g oxalic acid dihydrate is made up to 250 ml. The volume of 0.1 N NaOH required to completely neutralize 10 ml of this solution is (2001S)a)40 mlb)20 mlc)10 mld)4 mlCorrect answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about An aqueous solution of 6.3 g oxalic acid dihydrate is made up to 250 ml. The volume of 0.1 N NaOH required to completely neutralize 10 ml of this solution is (2001S)a)40 mlb)20 mlc)10 mld)4 mlCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An aqueous solution of 6.3 g oxalic acid dihydrate is made up to 250 ml. The volume of 0.1 N NaOH required to completely neutralize 10 ml of this solution is (2001S)a)40 mlb)20 mlc)10 mld)4 mlCorrect answer is option 'A'. Can you explain this answer?.
An aqueous solution of 6.3 g oxalic acid dihydrate is made up to 250 ml. The volume of 0.1 N NaOH required to completely neutralize 10 ml of this solution is (2001S)a)40 mlb)20 mlc)10 mld)4 mlCorrect answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about An aqueous solution of 6.3 g oxalic acid dihydrate is made up to 250 ml. The volume of 0.1 N NaOH required to completely neutralize 10 ml of this solution is (2001S)a)40 mlb)20 mlc)10 mld)4 mlCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An aqueous solution of 6.3 g oxalic acid dihydrate is made up to 250 ml. The volume of 0.1 N NaOH required to completely neutralize 10 ml of this solution is (2001S)a)40 mlb)20 mlc)10 mld)4 mlCorrect answer is option 'A'. Can you explain this answer?.
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