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12.5 ml of a solution, containing 6.0 g of a dibasic acid in one litre, was found to be neutralized by 10 ml of decinormal solution of NaOH. The molecular weight of the acid is
- a)110
- b)120
- c)115
- d)75
Correct answer is option 'D'. Can you explain this answer?
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12.5 ml of a solution, containing 6.0 g of a dibasic acid in one litre...
Solution:
Given, 12.5 ml of a solution containing 6.0 g of a dibasic acid in one litre is neutralized by 10 ml of decinormal solution of NaOH.
We need to find the molecular weight of the acid.
Let's first calculate the molarity of the NaOH solution.
1 L of decinormal solution of NaOH = 0.1 N NaOH = 40 g NaOH
∴ 1 L of decinormal solution of NaOH contains 40 g NaOH
Thus, 10 ml of decinormal solution of NaOH contains (40/1000) × 10 = 0.4 g NaOH
Now, let's find out the number of moles of NaOH used in the neutralization reaction.
Moles of NaOH = Mass of NaOH/Molar mass of NaOH
= 0.4/40
= 0.01 mol
Since the acid is dibasic, it will require two moles of NaOH for complete neutralization.
Thus, number of moles of dibasic acid in 12.5 ml of the solution = 0.01/2 = 0.005 mol
Now, let's calculate the molarity of the dibasic acid solution.
Molarity of dibasic acid solution = Number of moles of dibasic acid/Volume of solution in litres
= 0.005/(12.5/1000)
= 0.4 M
The formula for molarity is given as:
Molarity = Mass of solute (in g)/Molecular weight of solute × Volume of solution (in L)
On rearranging the formula, we get:
Molecular weight of solute = Mass of solute (in g)/Molarity × Volume of solution (in L)
Substituting the given values, we get:
Molecular weight of acid = 6/(0.4 × 1)
= 15 g/mol
Therefore, the molecular weight of the acid is 75 (2 × 15) g/mol.
Hence, option D - 75 is the correct answer.
Given, 12.5 ml of a solution containing 6.0 g of a dibasic acid in one litre is neutralized by 10 ml of decinormal solution of NaOH.
We need to find the molecular weight of the acid.
Let's first calculate the molarity of the NaOH solution.
1 L of decinormal solution of NaOH = 0.1 N NaOH = 40 g NaOH
∴ 1 L of decinormal solution of NaOH contains 40 g NaOH
Thus, 10 ml of decinormal solution of NaOH contains (40/1000) × 10 = 0.4 g NaOH
Now, let's find out the number of moles of NaOH used in the neutralization reaction.
Moles of NaOH = Mass of NaOH/Molar mass of NaOH
= 0.4/40
= 0.01 mol
Since the acid is dibasic, it will require two moles of NaOH for complete neutralization.
Thus, number of moles of dibasic acid in 12.5 ml of the solution = 0.01/2 = 0.005 mol
Now, let's calculate the molarity of the dibasic acid solution.
Molarity of dibasic acid solution = Number of moles of dibasic acid/Volume of solution in litres
= 0.005/(12.5/1000)
= 0.4 M
The formula for molarity is given as:
Molarity = Mass of solute (in g)/Molecular weight of solute × Volume of solution (in L)
On rearranging the formula, we get:
Molecular weight of solute = Mass of solute (in g)/Molarity × Volume of solution (in L)
Substituting the given values, we get:
Molecular weight of acid = 6/(0.4 × 1)
= 15 g/mol
Therefore, the molecular weight of the acid is 75 (2 × 15) g/mol.
Hence, option D - 75 is the correct answer.
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12.5 ml of a solution, containing 6.0 g of a dibasic acid in one litre, was found to be neutralized by 10 ml of decinormal solution of NaOH. The molecular weight of the acid isa)110b)120c)115d)75Correct answer is option 'D'. Can you explain this answer?
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12.5 ml of a solution, containing 6.0 g of a dibasic acid in one litre, was found to be neutralized by 10 ml of decinormal solution of NaOH. The molecular weight of the acid isa)110b)120c)115d)75Correct answer is option 'D'. 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 12.5 ml of a solution, containing 6.0 g of a dibasic acid in one litre, was found to be neutralized by 10 ml of decinormal solution of NaOH. The molecular weight of the acid isa)110b)120c)115d)75Correct answer is option 'D'. 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 12.5 ml of a solution, containing 6.0 g of a dibasic acid in one litre, was found to be neutralized by 10 ml of decinormal solution of NaOH. The molecular weight of the acid isa)110b)120c)115d)75Correct answer is option 'D'. Can you explain this answer?.
12.5 ml of a solution, containing 6.0 g of a dibasic acid in one litre, was found to be neutralized by 10 ml of decinormal solution of NaOH. The molecular weight of the acid isa)110b)120c)115d)75Correct answer is option 'D'. 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 12.5 ml of a solution, containing 6.0 g of a dibasic acid in one litre, was found to be neutralized by 10 ml of decinormal solution of NaOH. The molecular weight of the acid isa)110b)120c)115d)75Correct answer is option 'D'. 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 12.5 ml of a solution, containing 6.0 g of a dibasic acid in one litre, was found to be neutralized by 10 ml of decinormal solution of NaOH. The molecular weight of the acid isa)110b)120c)115d)75Correct answer is option 'D'. Can you explain this answer?.
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