JEE Exam  >  JEE Questions  >  5 ml of N HCl, 20 ml of N/2 H2SO4 and 30 ml o... Start Learning for Free
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 0.025
  • b)
    The normality of resulting mixture is 0.45
  • c)
    100 ml of this mixture is not neutralized completely by 50 ml of 0.050 N NaOH
  • d)
     100 ml of this mixture is neutralized completely by 50 ml of 0.025 M Ca(OH)2 solution
Correct answer is option 'A,D'. Can you explain this answer?
Verified Answer
5 ml of N HCl, 20 ml of N/2 H2SO4 and 30 ml of N/3 HNO3 are mixed toge...
As we know M(eq) = N x V
where
M(eq) = equivalent molarity
N = normality of solution
V = volume of solution
Applying this formula for each we get
For HCl----------M(eq) = 1 x 5 = 5M
For H₂SO₄---------M(eq) =  1/2 x 20 = 10M
For HNO₃----------M(eq) = 1/3 x 30 = 10M
For total solution, M(eq) = M₁ + M₂ + M₃
M(eq) = 5 + 10 + 10 = 25M
and volume = 1000 mL
Applying same formula M(eq) = N x V
25 = N x 1000
N = 25 / 1000
Normality = 0.04 N
View all questions of this test
Most Upvoted Answer
5 ml of N HCl, 20 ml of N/2 H2SO4 and 30 ml of N/3 HNO3 are mixed toge...
Solution:

Given:
Volume of N HCl = 5 ml
Volume of N/2 H2SO4 = 20 ml
Volume of N/3 HNO3 = 30 ml
Total volume = 1 L

Step 1: Calculate the number of moles of each acid

Number of moles of N HCl = N×V = 1×5/1000 = 0.005 mol
Number of moles of N/2 H2SO4 = N/2 × V = 1/2 × 20/1000 = 0.01 mol
Number of moles of N/3 HNO3 = N/3 × V = 1/3 × 30/1000 = 0.01 mol

Step 2: Calculate the total number of moles of acid

Total number of moles of acid = 0.005 + 0.01 + 0.01 = 0.025 mol

Step 3: Calculate the normality of the resulting mixture

Normality of the resulting mixture = Total number of moles of acid / Volume of mixture in L
= 0.025 / 1
= 0.025 N

Therefore, the correct option is (a).

Step 4: Calculate the number of moles of Ca(OH)2 required to neutralize the mixture

Number of moles of Ca(OH)2 required = Total number of moles of acid
= 0.025 mol

Step 5: Calculate the concentration of Ca(OH)2 solution required to neutralize the mixture

Volume of Ca(OH)2 solution required = Volume of mixture
= 100 ml
= 0.1 L

Concentration of Ca(OH)2 solution required = Number of moles of Ca(OH)2 required / Volume of Ca(OH)2 solution required
= 0.025 / 0.1
= 0.25 M

Step 6: Calculate the normality of Ca(OH)2 solution required to neutralize the mixture

Normality of Ca(OH)2 solution required = Concentration of Ca(OH)2 solution required / 2
= 0.25 / 2
= 0.125 N

Since the given Ca(OH)2 solution is 0.025 M or 0.0125 N, it will be able to completely neutralize the mixture. Therefore, the correct option is (d).
Explore Courses for JEE exam
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 0.025b)The normality of resulting mixture is 0.45c)100 ml of this mixture is not neutralized completely by 50 ml of 0.050 N NaOHd)100 ml of this mixture is neutralized completely by 50 ml of 0.025 M Ca(OH)2 solutionCorrect answer is option 'A,D'. Can you explain this answer?
Question Description
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 0.025b)The normality of resulting mixture is 0.45c)100 ml of this mixture is not neutralized completely by 50 ml of 0.050 N NaOHd)100 ml of this mixture is neutralized completely by 50 ml of 0.025 M Ca(OH)2 solutionCorrect 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 0.025b)The normality of resulting mixture is 0.45c)100 ml of this mixture is not neutralized completely by 50 ml of 0.050 N NaOHd)100 ml of this mixture is neutralized completely by 50 ml of 0.025 M Ca(OH)2 solutionCorrect 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 0.025b)The normality of resulting mixture is 0.45c)100 ml of this mixture is not neutralized completely by 50 ml of 0.050 N NaOHd)100 ml of this mixture is neutralized completely by 50 ml of 0.025 M Ca(OH)2 solutionCorrect 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 0.025b)The normality of resulting mixture is 0.45c)100 ml of this mixture is not neutralized completely by 50 ml of 0.050 N NaOHd)100 ml of this mixture is neutralized completely by 50 ml of 0.025 M Ca(OH)2 solutionCorrect answer is option 'A,D'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
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 0.025b)The normality of resulting mixture is 0.45c)100 ml of this mixture is not neutralized completely by 50 ml of 0.050 N NaOHd)100 ml of this mixture is neutralized completely by 50 ml of 0.025 M Ca(OH)2 solutionCorrect 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 0.025b)The normality of resulting mixture is 0.45c)100 ml of this mixture is not neutralized completely by 50 ml of 0.050 N NaOHd)100 ml of this mixture is neutralized completely by 50 ml of 0.025 M Ca(OH)2 solutionCorrect 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 0.025b)The normality of resulting mixture is 0.45c)100 ml of this mixture is not neutralized completely by 50 ml of 0.050 N NaOHd)100 ml of this mixture is neutralized completely by 50 ml of 0.025 M Ca(OH)2 solutionCorrect 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 0.025b)The normality of resulting mixture is 0.45c)100 ml of this mixture is not neutralized completely by 50 ml of 0.050 N NaOHd)100 ml of this mixture is neutralized completely by 50 ml of 0.025 M Ca(OH)2 solutionCorrect 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 0.025b)The normality of resulting mixture is 0.45c)100 ml of this mixture is not neutralized completely by 50 ml of 0.050 N NaOHd)100 ml of this mixture is neutralized completely by 50 ml of 0.025 M Ca(OH)2 solutionCorrect answer is option 'A,D'. Can you explain this answer? tests, examples and also practice JEE tests.
Explore Courses for JEE exam

Top Courses for JEE

Explore Courses
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev