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0.52 g of a dibasic acid required 100 mL of 0.1 N NaOH for complete neutralisation. The equivalent weight of acid is:
- a)26
- b)52
- c)104
- d)156
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
0.52 g of a dibasic acid required 100 mL of 0.1 N NaOH for complete ne...
Meq. of acid = Meq. of NaOH
∴ E = 52
∴ E = 52
Most Upvoted Answer
0.52 g of a dibasic acid required 100 mL of 0.1 N NaOH for complete ne...
To find the equivalent weight of the dibasic acid, we need to use the concept of normality (N) and the equation:
Normality (N) = (Weight of acid in grams)/(Equivalent weight of acid)
We are given the following information:
- Volume of NaOH used = 100 mL = 0.1 L
- Normality of NaOH = 0.1 N
- Weight of acid used = 0.52 g
To find the equivalent weight of the acid, we need to calculate the normality of the acid, which can be done using the equation:
Normality of acid = Normality of base × Volume of base/Volume of acid
In this case, the normality of the acid and base will be the same, as they are in a 1:1 ratio. Therefore, we can write:
0.1 N (NaOH) = Normality of acid × 0.1 L/Volume of acid
Simplifying the equation gives:
Normality of acid = 0.1 N × (Volume of acid/0.1 L)
Substituting the given values:
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N =
Normality (N) = (Weight of acid in grams)/(Equivalent weight of acid)
We are given the following information:
- Volume of NaOH used = 100 mL = 0.1 L
- Normality of NaOH = 0.1 N
- Weight of acid used = 0.52 g
To find the equivalent weight of the acid, we need to calculate the normality of the acid, which can be done using the equation:
Normality of acid = Normality of base × Volume of base/Volume of acid
In this case, the normality of the acid and base will be the same, as they are in a 1:1 ratio. Therefore, we can write:
0.1 N (NaOH) = Normality of acid × 0.1 L/Volume of acid
Simplifying the equation gives:
Normality of acid = 0.1 N × (Volume of acid/0.1 L)
Substituting the given values:
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N = 0.1 N × (Volume of acid/0.1 L)
0.1 N =
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0.52 g of a dibasic acid required 100 mL of 0.1 N NaOH for complete neutralisation. The equivalent weight of acid is: a)26 b)52 c)104 d)156Correct answer is option 'B'. Can you explain this answer?
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0.52 g of a dibasic acid required 100 mL of 0.1 N NaOH for complete neutralisation. The equivalent weight of acid is: a)26 b)52 c)104 d)156Correct answer is option 'B'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about 0.52 g of a dibasic acid required 100 mL of 0.1 N NaOH for complete neutralisation. The equivalent weight of acid is: a)26 b)52 c)104 d)156Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 0.52 g of a dibasic acid required 100 mL of 0.1 N NaOH for complete neutralisation. The equivalent weight of acid is: a)26 b)52 c)104 d)156Correct answer is option 'B'. Can you explain this answer?.
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