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5 ml of N HCl, 20 ml of N/2 H2SO4 and 30 ml of N/3 HNO3 are mixed together and volume made to one litre. Then,
- a)The normality of resulting mixture is 0.45
- b)The normality of resulting mixture is 0.025
- c)100 ml of this mixture is neutralized completely by 50 ml of 0.025 M Ca(OH)2 solution
- d)100 ml of this mixture is not neutralized completely by 50 ml of 0.050 N NaOH
Correct answer is option 'A,D'. Can you explain this answer?
Most Upvoted Answer
5 ml of N HCl, 20 ml of N/2 H2SO4 and 30 ml of N/3 HNO3 are mixed toge...
To determine the normality of the resulting mixture, we need to calculate the total number of equivalents present in the solution.
Given:
Volume of N HCl = 5 ml
Volume of N/2 H2SO4 = 20 ml
Volume of N/3 HNO3 = 30 ml
Step 1: Calculate the number of equivalents of each acid.
For N HCl:
Normality (N) = Number of equivalents / Volume (L)
1N = Number of equivalents / 0.005 L
Number of equivalents = 1N x 0.005 L = 0.005 equivalents
For N/2 H2SO4:
Normality (N/2) = Number of equivalents / Volume (L)
N/2 = Number of equivalents / 0.02 L
Number of equivalents = (N/2) x 0.02 L = 0.01 equivalents
For N/3 HNO3:
Normality (N/3) = Number of equivalents / Volume (L)
N/3 = Number of equivalents / 0.03 L
Number of equivalents = (N/3) x 0.03 L = 0.01 equivalents
Step 2: Calculate the total number of equivalents in the mixture.
Total number of equivalents = Number of equivalents of HCl + Number of equivalents of H2SO4 + Number of equivalents of HNO3
Total number of equivalents = 0.005 equivalents + 0.01 equivalents + 0.01 equivalents
Total number of equivalents = 0.025 equivalents
Step 3: Calculate the normality of the resulting mixture.
Normality = Total number of equivalents / Volume (L)
Normality = 0.025 equivalents / 1 L
Normality = 0.025 N
Therefore, the normality of the resulting mixture is 0.025 N (option B).
Now, let's move on to the second part of the question.
Given:
Volume of mixture = 100 ml
Volume of 0.050 N NaOH = 50 ml
Step 1: Calculate the number of equivalents of NaOH.
Normality (N) = Number of equivalents / Volume (L)
0.050 N = Number of equivalents / 0.050 L
Number of equivalents = 0.050 N x 0.050 L = 0.0025 equivalents
Step 2: Calculate the number of equivalents in the mixture.
Number of equivalents in the mixture = Normality x Volume (L)
Number of equivalents in the mixture = 0.025 N x 0.100 L = 0.0025 equivalents
Step 3: Compare the number of equivalents of NaOH with the number of equivalents in the mixture.
Since the number of equivalents of NaOH (0.0025 equivalents) is equal to the number of equivalents in the mixture (0.0025 equivalents), it means that the mixture is completely neutralized by the NaOH solution.
Therefore, the statement "100 ml of this mixture is neutralized completely by 50 ml of 0.050 N NaOH solution" is correct (option C).
In conclusion, the correct answers to the given options are A and C.
Given:
Volume of N HCl = 5 ml
Volume of N/2 H2SO4 = 20 ml
Volume of N/3 HNO3 = 30 ml
Step 1: Calculate the number of equivalents of each acid.
For N HCl:
Normality (N) = Number of equivalents / Volume (L)
1N = Number of equivalents / 0.005 L
Number of equivalents = 1N x 0.005 L = 0.005 equivalents
For N/2 H2SO4:
Normality (N/2) = Number of equivalents / Volume (L)
N/2 = Number of equivalents / 0.02 L
Number of equivalents = (N/2) x 0.02 L = 0.01 equivalents
For N/3 HNO3:
Normality (N/3) = Number of equivalents / Volume (L)
N/3 = Number of equivalents / 0.03 L
Number of equivalents = (N/3) x 0.03 L = 0.01 equivalents
Step 2: Calculate the total number of equivalents in the mixture.
Total number of equivalents = Number of equivalents of HCl + Number of equivalents of H2SO4 + Number of equivalents of HNO3
Total number of equivalents = 0.005 equivalents + 0.01 equivalents + 0.01 equivalents
Total number of equivalents = 0.025 equivalents
Step 3: Calculate the normality of the resulting mixture.
Normality = Total number of equivalents / Volume (L)
Normality = 0.025 equivalents / 1 L
Normality = 0.025 N
Therefore, the normality of the resulting mixture is 0.025 N (option B).
Now, let's move on to the second part of the question.
Given:
Volume of mixture = 100 ml
Volume of 0.050 N NaOH = 50 ml
Step 1: Calculate the number of equivalents of NaOH.
Normality (N) = Number of equivalents / Volume (L)
0.050 N = Number of equivalents / 0.050 L
Number of equivalents = 0.050 N x 0.050 L = 0.0025 equivalents
Step 2: Calculate the number of equivalents in the mixture.
Number of equivalents in the mixture = Normality x Volume (L)
Number of equivalents in the mixture = 0.025 N x 0.100 L = 0.0025 equivalents
Step 3: Compare the number of equivalents of NaOH with the number of equivalents in the mixture.
Since the number of equivalents of NaOH (0.0025 equivalents) is equal to the number of equivalents in the mixture (0.0025 equivalents), it means that the mixture is completely neutralized by the NaOH solution.
Therefore, the statement "100 ml of this mixture is neutralized completely by 50 ml of 0.050 N NaOH solution" is correct (option C).
In conclusion, the correct answers to the given options are A and C.
Community Answer
5 ml of N HCl, 20 ml of N/2 H2SO4 and 30 ml of N/3 HNO3 are mixed toge...
A
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5 ml of N HCl, 20 ml of N/2 H2SO4 and 30 ml of N/3 HNO3 are mixed together and volume made to one litre. Then,a)The normality of resulting mixture is 0.45b)The normality of resulting mixture is 0.025c)100 ml of this mixture is neutralized completely by 50 ml of 0.025 M Ca(OH)2 solutiond)100 ml of this mixture is not neutralized completely by 50 ml of 0.050 N NaOHCorrect answer is option 'A,D'. Can you explain this answer?
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5 ml of N HCl, 20 ml of N/2 H2SO4 and 30 ml of N/3 HNO3 are mixed together and volume made to one litre. Then,a)The normality of resulting mixture is 0.45b)The normality of resulting mixture is 0.025c)100 ml of this mixture is neutralized completely by 50 ml of 0.025 M Ca(OH)2 solutiond)100 ml of this mixture is not neutralized completely by 50 ml of 0.050 N NaOHCorrect answer is option 'A,D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 5 ml of N HCl, 20 ml of N/2 H2SO4 and 30 ml of N/3 HNO3 are mixed together and volume made to one litre. Then,a)The normality of resulting mixture is 0.45b)The normality of resulting mixture is 0.025c)100 ml of this mixture is neutralized completely by 50 ml of 0.025 M Ca(OH)2 solutiond)100 ml of this mixture is not neutralized completely by 50 ml of 0.050 N NaOHCorrect answer is option 'A,D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 5 ml of N HCl, 20 ml of N/2 H2SO4 and 30 ml of N/3 HNO3 are mixed together and volume made to one litre. Then,a)The normality of resulting mixture is 0.45b)The normality of resulting mixture is 0.025c)100 ml of this mixture is neutralized completely by 50 ml of 0.025 M Ca(OH)2 solutiond)100 ml of this mixture is not neutralized completely by 50 ml of 0.050 N NaOHCorrect answer is option 'A,D'. Can you explain this answer?.
5 ml of N HCl, 20 ml of N/2 H2SO4 and 30 ml of N/3 HNO3 are mixed together and volume made to one litre. Then,a)The normality of resulting mixture is 0.45b)The normality of resulting mixture is 0.025c)100 ml of this mixture is neutralized completely by 50 ml of 0.025 M Ca(OH)2 solutiond)100 ml of this mixture is not neutralized completely by 50 ml of 0.050 N NaOHCorrect answer is option 'A,D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 5 ml of N HCl, 20 ml of N/2 H2SO4 and 30 ml of N/3 HNO3 are mixed together and volume made to one litre. Then,a)The normality of resulting mixture is 0.45b)The normality of resulting mixture is 0.025c)100 ml of this mixture is neutralized completely by 50 ml of 0.025 M Ca(OH)2 solutiond)100 ml of this mixture is not neutralized completely by 50 ml of 0.050 N NaOHCorrect answer is option 'A,D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 5 ml of N HCl, 20 ml of N/2 H2SO4 and 30 ml of N/3 HNO3 are mixed together and volume made to one litre. Then,a)The normality of resulting mixture is 0.45b)The normality of resulting mixture is 0.025c)100 ml of this mixture is neutralized completely by 50 ml of 0.025 M Ca(OH)2 solutiond)100 ml of this mixture is not neutralized completely by 50 ml of 0.050 N NaOHCorrect answer is option 'A,D'. Can you explain this answer?.
Solutions for 5 ml of N HCl, 20 ml of N/2 H2SO4 and 30 ml of N/3 HNO3 are mixed together and volume made to one litre. Then,a)The normality of resulting mixture is 0.45b)The normality of resulting mixture is 0.025c)100 ml of this mixture is neutralized completely by 50 ml of 0.025 M Ca(OH)2 solutiond)100 ml of this mixture is not neutralized completely by 50 ml of 0.050 N NaOHCorrect answer is option 'A,D'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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Here you can find the meaning of 5 ml of N HCl, 20 ml of N/2 H2SO4 and 30 ml of N/3 HNO3 are mixed together and volume made to one litre. Then,a)The normality of resulting mixture is 0.45b)The normality of resulting mixture is 0.025c)100 ml of this mixture is neutralized completely by 50 ml of 0.025 M Ca(OH)2 solutiond)100 ml of this mixture is not neutralized completely by 50 ml of 0.050 N NaOHCorrect answer is option 'A,D'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
5 ml of N HCl, 20 ml of N/2 H2SO4 and 30 ml of N/3 HNO3 are mixed together and volume made to one litre. Then,a)The normality of resulting mixture is 0.45b)The normality of resulting mixture is 0.025c)100 ml of this mixture is neutralized completely by 50 ml of 0.025 M Ca(OH)2 solutiond)100 ml of this mixture is not neutralized completely by 50 ml of 0.050 N NaOHCorrect answer is option 'A,D'. Can you explain this answer?, a detailed solution for 5 ml of N HCl, 20 ml of N/2 H2SO4 and 30 ml of N/3 HNO3 are mixed together and volume made to one litre. Then,a)The normality of resulting mixture is 0.45b)The normality of resulting mixture is 0.025c)100 ml of this mixture is neutralized completely by 50 ml of 0.025 M Ca(OH)2 solutiond)100 ml of this mixture is not neutralized completely by 50 ml of 0.050 N NaOHCorrect answer is option 'A,D'. Can you explain this answer? has been provided alongside types of 5 ml of N HCl, 20 ml of N/2 H2SO4 and 30 ml of N/3 HNO3 are mixed together and volume made to one litre. Then,a)The normality of resulting mixture is 0.45b)The normality of resulting mixture is 0.025c)100 ml of this mixture is neutralized completely by 50 ml of 0.025 M Ca(OH)2 solutiond)100 ml of this mixture is not neutralized completely by 50 ml of 0.050 N NaOHCorrect answer is option 'A,D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice 5 ml of N HCl, 20 ml of N/2 H2SO4 and 30 ml of N/3 HNO3 are mixed together and volume made to one litre. Then,a)The normality of resulting mixture is 0.45b)The normality of resulting mixture is 0.025c)100 ml of this mixture is neutralized completely by 50 ml of 0.025 M Ca(OH)2 solutiond)100 ml of this mixture is not neutralized completely by 50 ml of 0.050 N NaOHCorrect answer is option 'A,D'. Can you explain this answer? tests, examples and also practice JEE tests.
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