Class 12 Exam > Class 12 Questions > 0.52 g of a dibasic acid required 100 mL of 0...
Start Learning for Free
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?
| FREE This question is part of | Download PDF Attempt this Test |
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 =
|
Explore Courses for Class 12 exam
|
|
Similar Class 12 Doubts
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?
Question Description
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?.
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?.
Solutions 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? in English & in Hindi are available as part of our courses for Class 12.
Download more important topics, notes, lectures and mock test series for Class 12 Exam by signing up for free.
Here you can find the meaning of 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? defined & explained in the simplest way possible. Besides giving the explanation of
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?, a detailed solution 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? has been provided alongside types of 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? theory, EduRev gives you an
ample number of questions to practice 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? tests, examples and also practice Class 12 tests.
|
|
Explore Courses for Class 12 exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup