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12.5 gm of fuming H2SO4 (labelled as 112%) is mixed with 100 lit water. Molar concentration of H+ in resultant solution is :
[Note : Assume that H2SO4 dissociate completely and there is no change in volume on mixing]
[Note : Assume that H2SO4 dissociate completely and there is no change in volume on mixing]
- a)2/700
- b)2/350
- c)3/350
- d)3/700
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
12.5 gm of fuming H2SO4(labelled as 112%) is mixed with 100 lit water....
The correct answer is Option A.
Given :
mass = 12.5 gm
% of H2SO4 = 112 %
= 112 /100
mass = 12.5 gm
% of H2SO4 = 112 %
= 112 /100
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12.5 gm of fuming H2SO4(labelled as 112%) is mixed with 100 lit water....
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12.5 gm of fuming H2SO4(labelled as 112%) is mixed with 100 lit water....
Given:
Mass of fuming H2SO4 = 12.5 g
Concentration of fuming H2SO4 = 112%
Volume of water = 100 L
To find: Molar concentration of H in the resultant solution
Solution:
1. Calculate the number of moles of H2SO4 in the given mass:
Molar mass of H2SO4 = 2(1.008) + 32.06 + 4(16.00) = 98.08 g/mol
Number of moles of H2SO4 = (12.5 g) / (98.08 g/mol) = 0.1274 mol
2. Calculate the number of moles of H in the given mass of H2SO4:
Number of moles of H = 2 × 0.1274 mol = 0.2548 mol
3. Calculate the total number of moles of solute (H2SO4 and H) in the solution:
Total number of moles of solute = 0.1274 mol (from H2SO4) + 0.2548 mol (from H) = 0.3822 mol
4. Calculate the molarity of the solution:
Molarity = (Total number of moles of solute) / (Volume of solution in L)
Volume of solution = Volume of water = 100 L
Molarity = 0.3822 mol / 100 L = 0.003822 M
5. Calculate the molar concentration of H in the solution:
Molar concentration of H = (Number of moles of H) / (Volume of solution in L)
Molar concentration of H = 0.2548 mol / 100 L = 0.002548 M
Answer: The molar concentration of H in the resultant solution is 0.002548 M, which is approximately equal to 2/700. Therefore, the correct option is (a) 2/700.
Mass of fuming H2SO4 = 12.5 g
Concentration of fuming H2SO4 = 112%
Volume of water = 100 L
To find: Molar concentration of H in the resultant solution
Solution:
1. Calculate the number of moles of H2SO4 in the given mass:
Molar mass of H2SO4 = 2(1.008) + 32.06 + 4(16.00) = 98.08 g/mol
Number of moles of H2SO4 = (12.5 g) / (98.08 g/mol) = 0.1274 mol
2. Calculate the number of moles of H in the given mass of H2SO4:
Number of moles of H = 2 × 0.1274 mol = 0.2548 mol
3. Calculate the total number of moles of solute (H2SO4 and H) in the solution:
Total number of moles of solute = 0.1274 mol (from H2SO4) + 0.2548 mol (from H) = 0.3822 mol
4. Calculate the molarity of the solution:
Molarity = (Total number of moles of solute) / (Volume of solution in L)
Volume of solution = Volume of water = 100 L
Molarity = 0.3822 mol / 100 L = 0.003822 M
5. Calculate the molar concentration of H in the solution:
Molar concentration of H = (Number of moles of H) / (Volume of solution in L)
Molar concentration of H = 0.2548 mol / 100 L = 0.002548 M
Answer: The molar concentration of H in the resultant solution is 0.002548 M, which is approximately equal to 2/700. Therefore, the correct option is (a) 2/700.
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12.5 gm of fuming H2SO4(labelled as 112%) is mixed with 100 lit water. Molar concentration of H+ in resultant solution is :[Note : Assume that H2SO4dissociate completely and there is no change in volume on mixing]a)2/700b)2/350c)3/350d)3/700Correct answer is option 'A'. Can you explain this answer?
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12.5 gm of fuming H2SO4(labelled as 112%) is mixed with 100 lit water. Molar concentration of H+ in resultant solution is :[Note : Assume that H2SO4dissociate completely and there is no change in volume on mixing]a)2/700b)2/350c)3/350d)3/700Correct answer is option 'A'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about 12.5 gm of fuming H2SO4(labelled as 112%) is mixed with 100 lit water. Molar concentration of H+ in resultant solution is :[Note : Assume that H2SO4dissociate completely and there is no change in volume on mixing]a)2/700b)2/350c)3/350d)3/700Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 12.5 gm of fuming H2SO4(labelled as 112%) is mixed with 100 lit water. Molar concentration of H+ in resultant solution is :[Note : Assume that H2SO4dissociate completely and there is no change in volume on mixing]a)2/700b)2/350c)3/350d)3/700Correct answer is option 'A'. Can you explain this answer?.
12.5 gm of fuming H2SO4(labelled as 112%) is mixed with 100 lit water. Molar concentration of H+ in resultant solution is :[Note : Assume that H2SO4dissociate completely and there is no change in volume on mixing]a)2/700b)2/350c)3/350d)3/700Correct answer is option 'A'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about 12.5 gm of fuming H2SO4(labelled as 112%) is mixed with 100 lit water. Molar concentration of H+ in resultant solution is :[Note : Assume that H2SO4dissociate completely and there is no change in volume on mixing]a)2/700b)2/350c)3/350d)3/700Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 12.5 gm of fuming H2SO4(labelled as 112%) is mixed with 100 lit water. Molar concentration of H+ in resultant solution is :[Note : Assume that H2SO4dissociate completely and there is no change in volume on mixing]a)2/700b)2/350c)3/350d)3/700Correct answer is option 'A'. Can you explain this answer?.
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