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30 ml of solution is neutralised by 15 ml of 0.2 N base. The strength of the acid solution is
  • a)
    0.1 N
  • b)
    0.15 N
  • c)
    0.3 N
  • d)
    0.4 N
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
30 ml of solution is neutralised by 15 ml of 0.2 N base. The strength ...
To determine the strength of the acid solution, follow these steps:
  • The normality of the base is given as 0.2 N.
  • The volume of the base used is 15 ml.
  • Calculate the equivalents of the base:
    • Equivalents = Normality × Volume = 0.2 N × 15 ml = 3 equivalents.
  • The volume of the acid solution is 30 ml.
  • Since the base neutralises the acid, the equivalents of the acid are also 3.
  • Now, calculate the normality of the acid:
    • Normality = Equivalents / Volume = 3 equivalents / 30 ml = 0.1 N.
Thus, the strength of the acid solution is 0.1 N.
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Community Answer
30 ml of solution is neutralised by 15 ml of 0.2 N base. The strength ...
Use, N1*v1=N2*v2
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30 ml of solution is neutralised by 15 ml of 0.2 N base. The strength of the acid solution isa)0.1 Nb)0.15 Nc)0.3 Nd)0.4 NCorrect answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 30 ml of solution is neutralised by 15 ml of 0.2 N base. The strength of the acid solution isa)0.1 Nb)0.15 Nc)0.3 Nd)0.4 NCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 30 ml of solution is neutralised by 15 ml of 0.2 N base. The strength of the acid solution isa)0.1 Nb)0.15 Nc)0.3 Nd)0.4 NCorrect answer is option 'A'. Can you explain this answer?.
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