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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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