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0.6 g of urea on strong heating with NaOH evolves NH3. Liberated NH3 will combine completely with which of the following HCl solution ?
- a)100 mL of 0.2 N HCl
- b)400 mL of 0.2 N HCl
- c)100 mL of 0.1 N HCl
- d)200 mL of 0.2 N HCl
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
0.6 g of urea on strong heating with NaOH evolves NH3. Liberated NH3 w...
2 � mole of Urea ≡ mole of NH3 ....(1)
mole of NH3 ≡ mole of HCl ....(2)
∴ mole of HCl = 0.02 mole.
Most Upvoted Answer
0.6 g of urea on strong heating with NaOH evolves NH3. Liberated NH3 w...
Given:
- Mass of urea (NH2CONH2) = 0.6 g
- Urea on strong heating with NaOH evolves NH3
- We need to determine which HCl solution will completely react with the liberated NH3
Solution:
To solve this problem, we can use the concept of stoichiometry and the balanced chemical equation for the reaction between NH3 and HCl. The balanced chemical equation is as follows:
NH3 + HCl → NH4Cl
From the equation, we can see that one molecule of NH3 reacts with one molecule of HCl to form one molecule of NH4Cl. Therefore, the molar ratio between NH3 and HCl is 1:1.
To determine which HCl solution will completely react with the liberated NH3, we need to calculate the number of moles of NH3 and compare it with the number of moles of HCl in each solution.
Calculations:
1. Number of moles of urea:
- The molar mass of urea (NH2CONH2) = 14.01 + 2(1.01) + 12.01 + 2(16.00) = 60.06 g/mol
- Number of moles of urea = mass of urea / molar mass of urea = 0.6 g / 60.06 g/mol = 0.00999 mol (approximately 0.01 mol)
2. Number of moles of NH3:
- From the balanced chemical equation, we know that one mole of urea produces one mole of NH3.
- Therefore, the number of moles of NH3 = number of moles of urea = 0.01 mol
3. Comparing with the number of moles of HCl in each solution:
a) 100 mL of 0.2 N HCl:
- Number of moles of HCl = volume of HCl (in L) × molarity of HCl
- Volume of HCl = 100 mL = 100/1000 L = 0.1 L
- Number of moles of HCl = 0.1 L × 0.2 N = 0.02 mol
- Since the number of moles of NH3 (0.01 mol) is less than the number of moles of HCl (0.02 mol), the NH3 will not completely react with 100 mL of 0.2 N HCl.
b) 400 mL of 0.2 N HCl:
- Number of moles of HCl = volume of HCl (in L) × molarity of HCl
- Volume of HCl = 400 mL = 400/1000 L = 0.4 L
- Number of moles of HCl = 0.4 L × 0.2 N = 0.08 mol
- Since the number of moles of NH3 (0.01 mol) is less than the number of moles of HCl (0.08 mol), the NH3 will not completely react with 400 mL of 0.2 N HCl.
c) 100 mL of 0.1 N HCl:
- Mass of urea (NH2CONH2) = 0.6 g
- Urea on strong heating with NaOH evolves NH3
- We need to determine which HCl solution will completely react with the liberated NH3
Solution:
To solve this problem, we can use the concept of stoichiometry and the balanced chemical equation for the reaction between NH3 and HCl. The balanced chemical equation is as follows:
NH3 + HCl → NH4Cl
From the equation, we can see that one molecule of NH3 reacts with one molecule of HCl to form one molecule of NH4Cl. Therefore, the molar ratio between NH3 and HCl is 1:1.
To determine which HCl solution will completely react with the liberated NH3, we need to calculate the number of moles of NH3 and compare it with the number of moles of HCl in each solution.
Calculations:
1. Number of moles of urea:
- The molar mass of urea (NH2CONH2) = 14.01 + 2(1.01) + 12.01 + 2(16.00) = 60.06 g/mol
- Number of moles of urea = mass of urea / molar mass of urea = 0.6 g / 60.06 g/mol = 0.00999 mol (approximately 0.01 mol)
2. Number of moles of NH3:
- From the balanced chemical equation, we know that one mole of urea produces one mole of NH3.
- Therefore, the number of moles of NH3 = number of moles of urea = 0.01 mol
3. Comparing with the number of moles of HCl in each solution:
a) 100 mL of 0.2 N HCl:
- Number of moles of HCl = volume of HCl (in L) × molarity of HCl
- Volume of HCl = 100 mL = 100/1000 L = 0.1 L
- Number of moles of HCl = 0.1 L × 0.2 N = 0.02 mol
- Since the number of moles of NH3 (0.01 mol) is less than the number of moles of HCl (0.02 mol), the NH3 will not completely react with 100 mL of 0.2 N HCl.
b) 400 mL of 0.2 N HCl:
- Number of moles of HCl = volume of HCl (in L) × molarity of HCl
- Volume of HCl = 400 mL = 400/1000 L = 0.4 L
- Number of moles of HCl = 0.4 L × 0.2 N = 0.08 mol
- Since the number of moles of NH3 (0.01 mol) is less than the number of moles of HCl (0.08 mol), the NH3 will not completely react with 400 mL of 0.2 N HCl.
c) 100 mL of 0.1 N HCl:
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0.6 g of urea on strong heating with NaOH evolves NH3. Liberated NH3 will combine completely with which of the following HCl solution ?a)100 mL of 0.2 N HClb)400 mL of 0.2 N HClc)100 mL of 0.1 N HCld)200 mL of 0.2 N HClCorrect answer is option 'A'. Can you explain this answer?
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0.6 g of urea on strong heating with NaOH evolves NH3. Liberated NH3 will combine completely with which of the following HCl solution ?a)100 mL of 0.2 N HClb)400 mL of 0.2 N HClc)100 mL of 0.1 N HCld)200 mL of 0.2 N HClCorrect 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 0.6 g of urea on strong heating with NaOH evolves NH3. Liberated NH3 will combine completely with which of the following HCl solution ?a)100 mL of 0.2 N HClb)400 mL of 0.2 N HClc)100 mL of 0.1 N HCld)200 mL of 0.2 N HClCorrect 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 0.6 g of urea on strong heating with NaOH evolves NH3. Liberated NH3 will combine completely with which of the following HCl solution ?a)100 mL of 0.2 N HClb)400 mL of 0.2 N HClc)100 mL of 0.1 N HCld)200 mL of 0.2 N HClCorrect answer is option 'A'. Can you explain this answer?.
0.6 g of urea on strong heating with NaOH evolves NH3. Liberated NH3 will combine completely with which of the following HCl solution ?a)100 mL of 0.2 N HClb)400 mL of 0.2 N HClc)100 mL of 0.1 N HCld)200 mL of 0.2 N HClCorrect 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 0.6 g of urea on strong heating with NaOH evolves NH3. Liberated NH3 will combine completely with which of the following HCl solution ?a)100 mL of 0.2 N HClb)400 mL of 0.2 N HClc)100 mL of 0.1 N HCld)200 mL of 0.2 N HClCorrect 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 0.6 g of urea on strong heating with NaOH evolves NH3. Liberated NH3 will combine completely with which of the following HCl solution ?a)100 mL of 0.2 N HClb)400 mL of 0.2 N HClc)100 mL of 0.1 N HCld)200 mL of 0.2 N HClCorrect answer is option 'A'. Can you explain this answer?.
Solutions for 0.6 g of urea on strong heating with NaOH evolves NH3. Liberated NH3 will combine completely with which of the following HCl solution ?a)100 mL of 0.2 N HClb)400 mL of 0.2 N HClc)100 mL of 0.1 N HCld)200 mL of 0.2 N HClCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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0.6 g of urea on strong heating with NaOH evolves NH3. Liberated NH3 will combine completely with which of the following HCl solution ?a)100 mL of 0.2 N HClb)400 mL of 0.2 N HClc)100 mL of 0.1 N HCld)200 mL of 0.2 N HClCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for 0.6 g of urea on strong heating with NaOH evolves NH3. Liberated NH3 will combine completely with which of the following HCl solution ?a)100 mL of 0.2 N HClb)400 mL of 0.2 N HClc)100 mL of 0.1 N HCld)200 mL of 0.2 N HClCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of 0.6 g of urea on strong heating with NaOH evolves NH3. Liberated NH3 will combine completely with which of the following HCl solution ?a)100 mL of 0.2 N HClb)400 mL of 0.2 N HClc)100 mL of 0.1 N HCld)200 mL of 0.2 N HClCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
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