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20 ml of decinormal solution of NaOH neutralises 25 ml of a solution of dibasic acid containing 3g. of the acid per 500 ml. The Molecular weight of the acid is
- a)150
- b)75
- c)225
- d)300
Correct answer is option 'A'. Can you explain this answer?
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Given:
Volume of NaOH solution = 20 ml
Normality of NaOH solution = 0.1 N
Volume of dibasic acid solution = 25 ml
Concentration of dibasic acid solution = 3 g/500 ml
To find:
Molecular weight of the acid
Solution:
1. Calculation of number of moles of NaOH used:
Molarity of NaOH = 0.1 N
Volume of NaOH used = 20 ml = 0.02 L
Number of moles of NaOH used = Molarity x Volume = 0.1 x 0.02 = 0.002 mol
2. Calculation of number of moles of dibasic acid:
Volume of dibasic acid used = 25 ml = 0.025 L
Concentration of dibasic acid = 3 g/500 ml = 0.006 g/ml
Number of moles of dibasic acid = Mass/ Molecular weight
Mass of dibasic acid = Volume x Concentration = 0.025 x 0.006 = 0.00015 g
Number of moles of dibasic acid = 0.00015/Molecular weight
3. As per the reaction between NaOH and dibasic acid, 1 mole of NaOH reacts with 2 moles of dibasic acid.
Thus, 0.002 moles of NaOH will react with 2 x 0.00015 moles of dibasic acid.
4. Equating the number of moles of NaOH used and dibasic acid used:
0.002 = 2 x 0.00015/Molecular weight
Molecular weight = 150 g/mol
Therefore, the molecular weight of the dibasic acid is 150 g/mol, which is option (a).
Volume of NaOH solution = 20 ml
Normality of NaOH solution = 0.1 N
Volume of dibasic acid solution = 25 ml
Concentration of dibasic acid solution = 3 g/500 ml
To find:
Molecular weight of the acid
Solution:
1. Calculation of number of moles of NaOH used:
Molarity of NaOH = 0.1 N
Volume of NaOH used = 20 ml = 0.02 L
Number of moles of NaOH used = Molarity x Volume = 0.1 x 0.02 = 0.002 mol
2. Calculation of number of moles of dibasic acid:
Volume of dibasic acid used = 25 ml = 0.025 L
Concentration of dibasic acid = 3 g/500 ml = 0.006 g/ml
Number of moles of dibasic acid = Mass/ Molecular weight
Mass of dibasic acid = Volume x Concentration = 0.025 x 0.006 = 0.00015 g
Number of moles of dibasic acid = 0.00015/Molecular weight
3. As per the reaction between NaOH and dibasic acid, 1 mole of NaOH reacts with 2 moles of dibasic acid.
Thus, 0.002 moles of NaOH will react with 2 x 0.00015 moles of dibasic acid.
4. Equating the number of moles of NaOH used and dibasic acid used:
0.002 = 2 x 0.00015/Molecular weight
Molecular weight = 150 g/mol
Therefore, the molecular weight of the dibasic acid is 150 g/mol, which is option (a).
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20 ml of decinormal solution of NaOH neutralises 25 ml of a solution o...
20 ml multiply by deci normal solution means 0.1=volume of acid multiply by no. of gram equivalent of solute and just convert 500 ml in litres
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20 ml of decinormal solution of NaOH neutralises 25 ml of a solution of dibasic acid containing 3g. of the acid per 500 ml. The Molecular weight of the acid isa)150b)75c)225d)300Correct answer is option 'A'. Can you explain this answer?
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20 ml of decinormal solution of NaOH neutralises 25 ml of a solution of dibasic acid containing 3g. of the acid per 500 ml. The Molecular weight of the acid isa)150b)75c)225d)300Correct answer is option 'A'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about 20 ml of decinormal solution of NaOH neutralises 25 ml of a solution of dibasic acid containing 3g. of the acid per 500 ml. The Molecular weight of the acid isa)150b)75c)225d)300Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 20 ml of decinormal solution of NaOH neutralises 25 ml of a solution of dibasic acid containing 3g. of the acid per 500 ml. The Molecular weight of the acid isa)150b)75c)225d)300Correct answer is option 'A'. Can you explain this answer?.
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