12.5 mL of a solution containing 6g of a dibasic acid in 1L was found to be neutralised by 10mL of a decinormal solution of NaOH. The molecular mass of the acid is:
(a110
(b)75
(c)120
(d)150
Answer
Verified
123k+ views
Hint:Neutralisation is the process where a known quantity of acid and a base react to form salt and water. Using stoichiometric calculation we can solve the question.
Complete step by step answer
It is given that the amount of acid in 1L solution = 6g
So, the amount of acid in 12.5mL solution will be = \[\dfrac{6}{1000}\times 12.5g=0.075g\]
When it is at neutralisation,
10mL of \[\dfrac{N}{10}\]NaOH = 10mL of \[\dfrac{N}{10}\]acid (it is given that decinormal which means\[\dfrac{N}{10}\])
We calculated that \[\dfrac{N}{10}\]acid=0.075g
So, for 1000mL \[\dfrac{N}{10}\]acid=\[\dfrac{0.075}{10}\times 1000=7.5\]g
1000mL of N acid = 7.5x10 =75g
Therefore, we get the equivalent mass of the acid= 75
As the acid is dibasic, we need to multiply equivalent mass with 2 in order to calculate molecular mass.
Molecular mass= 75x2=150g
Thus, the correct answer to the question is option (d).
Additional Information:
Neutralisation reactions can be used for quantitative analysis of compounds. Either acid or base quantity would be known, that is it is fixed and we can find the quantity of the other. This question is one example for such analysis. It has many applications in industry and laboratories. In wastewater treatment, chemical neutralisation methods are used to decrease the damage caused by an effluent upon its release into the environment.
Note:
Only in neutralisation reaction we can take the condition 10mL of \[\dfrac{N}{10}\]NaOH = 10mL of \[\dfrac{N}{10}\]acid. We should also keep in mind to multiply the equivalent mass with 2, and then only we get the molecular mass.
Complete step by step answer
It is given that the amount of acid in 1L solution = 6g
So, the amount of acid in 12.5mL solution will be = \[\dfrac{6}{1000}\times 12.5g=0.075g\]
When it is at neutralisation,
10mL of \[\dfrac{N}{10}\]NaOH = 10mL of \[\dfrac{N}{10}\]acid (it is given that decinormal which means\[\dfrac{N}{10}\])
We calculated that \[\dfrac{N}{10}\]acid=0.075g
So, for 1000mL \[\dfrac{N}{10}\]acid=\[\dfrac{0.075}{10}\times 1000=7.5\]g
1000mL of N acid = 7.5x10 =75g
Therefore, we get the equivalent mass of the acid= 75
As the acid is dibasic, we need to multiply equivalent mass with 2 in order to calculate molecular mass.
Molecular mass= 75x2=150g
Thus, the correct answer to the question is option (d).
Additional Information:
Neutralisation reactions can be used for quantitative analysis of compounds. Either acid or base quantity would be known, that is it is fixed and we can find the quantity of the other. This question is one example for such analysis. It has many applications in industry and laboratories. In wastewater treatment, chemical neutralisation methods are used to decrease the damage caused by an effluent upon its release into the environment.
Note:
Only in neutralisation reaction we can take the condition 10mL of \[\dfrac{N}{10}\]NaOH = 10mL of \[\dfrac{N}{10}\]acid. We should also keep in mind to multiply the equivalent mass with 2, and then only we get the molecular mass.
Recently Updated Pages
The hybridization and shape of NH2 ion are a sp2 and class 11 chemistry JEE_Main
Total number of orbitals associated with the 3rd shell class 11 chemistry JEE_Main
Which of the following has the lowest boiling point class 11 chemistry JEE_Main
Which of the following compounds has zero dipole moment class 11 chemistry JEE_Main
Number of g of oxygen in 322 g Na2SO410H2O is Molwt class 11 chemistry JEE_Main
In the neutralization process of H3PO4 and NaOH the class 11 chemistry JEE_Main
Trending doubts
JEE Mains 2025: Check Important Dates, Syllabus, Exam Pattern, Fee and Updates
JEE Main Login 2045: Step-by-Step Instructions and Details
JEE Main Chemistry Question Paper with Answer Keys and Solutions
JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking
JEE Main 2023 January 24 Shift 2 Question Paper with Answer Keys & Solutions
JEE Main Chemistry Exam Pattern 2025
Other Pages
NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reaction
NCERT Solutions for Class 11 Chemistry Chapter 5 Thermodynamics
NCERT Solutions for Class 11 Chemistry Chapter 8 Organic Chemistry
NCERT Solutions for Class 11 Chemistry Chapter 6 Equilibrium
NCERT Solutions for Class 11 Chemistry Chapter 9 Hydrocarbons
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs