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An excess of AgNO3 is added to 100 mL of a 0.01 M solution of dichlorotetraaquachromium (iii) chloride. The number of moles of AgCl precipitated would be : [NEET 2013]
- a)0.002
- b)0.003
- c)0.01
- d)0.001
Correct answer is option 'D'. Can you explain this answer?
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An excess of AgNO3 is added to 100 mL of a 0.01 M solution of dichloro...
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To solve this problem, we need to determine the number of moles of AgCl that will be precipitated when an excess of AgNO3 is added to a solution containing dichlorotetraaquachromium (III) chloride.
Before we proceed with the calculation, let's first write the balanced chemical equation for the reaction:
2 AgNO3 + CrCl3.4H2O → AgCl + Cr(NO3)3.4H2O
From the balanced equation, we can see that 2 moles of AgNO3 react with 1 mole of CrCl3 to form 1 mole of AgCl.
Now, let's calculate the number of moles of CrCl3 in the given solution:
Molarity (M) = Moles (n) / Volume (V)
n = M x V
Given:
Molarity of CrCl3 solution = 0.01 M
Volume of CrCl3 solution = 100 mL = 0.1 L
n(CrCl3) = 0.01 M x 0.1 L = 0.001 moles
Since the stoichiometry of the reaction tells us that 2 moles of AgNO3 react with 1 mole of CrCl3, we can conclude that 0.002 moles of AgNO3 are required to react with 0.001 moles of CrCl3.
However, the problem states that an excess of AgNO3 is added, which means that there is more than enough AgNO3 to react with all the CrCl3 present.
Therefore, the number of moles of AgCl precipitated will be equal to the number of moles of CrCl3 present in the solution, which is 0.001 moles.
Hence, the correct answer is option D) 0.001.
Before we proceed with the calculation, let's first write the balanced chemical equation for the reaction:
2 AgNO3 + CrCl3.4H2O → AgCl + Cr(NO3)3.4H2O
From the balanced equation, we can see that 2 moles of AgNO3 react with 1 mole of CrCl3 to form 1 mole of AgCl.
Now, let's calculate the number of moles of CrCl3 in the given solution:
Molarity (M) = Moles (n) / Volume (V)
n = M x V
Given:
Molarity of CrCl3 solution = 0.01 M
Volume of CrCl3 solution = 100 mL = 0.1 L
n(CrCl3) = 0.01 M x 0.1 L = 0.001 moles
Since the stoichiometry of the reaction tells us that 2 moles of AgNO3 react with 1 mole of CrCl3, we can conclude that 0.002 moles of AgNO3 are required to react with 0.001 moles of CrCl3.
However, the problem states that an excess of AgNO3 is added, which means that there is more than enough AgNO3 to react with all the CrCl3 present.
Therefore, the number of moles of AgCl precipitated will be equal to the number of moles of CrCl3 present in the solution, which is 0.001 moles.
Hence, the correct answer is option D) 0.001.
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An excess of AgNO3 is added to 100 mL of a 0.01 M solution of dichlorotetraaquachromium (iii) chloride. The number of moles of AgCl precipitated would be : [NEET 2013]a)0.002b)0.003c)0.01d)0.001Correct answer is option 'D'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about An excess of AgNO3 is added to 100 mL of a 0.01 M solution of dichlorotetraaquachromium (iii) chloride. The number of moles of AgCl precipitated would be : [NEET 2013]a)0.002b)0.003c)0.01d)0.001Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An excess of AgNO3 is added to 100 mL of a 0.01 M solution of dichlorotetraaquachromium (iii) chloride. The number of moles of AgCl precipitated would be : [NEET 2013]a)0.002b)0.003c)0.01d)0.001Correct answer is option 'D'. Can you explain this answer?.
An excess of AgNO3 is added to 100 mL of a 0.01 M solution of dichlorotetraaquachromium (iii) chloride. The number of moles of AgCl precipitated would be : [NEET 2013]a)0.002b)0.003c)0.01d)0.001Correct answer is option 'D'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about An excess of AgNO3 is added to 100 mL of a 0.01 M solution of dichlorotetraaquachromium (iii) chloride. The number of moles of AgCl precipitated would be : [NEET 2013]a)0.002b)0.003c)0.01d)0.001Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An excess of AgNO3 is added to 100 mL of a 0.01 M solution of dichlorotetraaquachromium (iii) chloride. The number of moles of AgCl precipitated would be : [NEET 2013]a)0.002b)0.003c)0.01d)0.001Correct answer is option 'D'. Can you explain this answer?.
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An excess of AgNO3 is added to 100 mL of a 0.01 M solution of dichlorotetraaquachromium (iii) chloride. The number of moles of AgCl precipitated would be : [NEET 2013]a)0.002b)0.003c)0.01d)0.001Correct answer is option 'D'. Can you explain this answer?, a detailed solution for An excess of AgNO3 is added to 100 mL of a 0.01 M solution of dichlorotetraaquachromium (iii) chloride. The number of moles of AgCl precipitated would be : [NEET 2013]a)0.002b)0.003c)0.01d)0.001Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of An excess of AgNO3 is added to 100 mL of a 0.01 M solution of dichlorotetraaquachromium (iii) chloride. The number of moles of AgCl precipitated would be : [NEET 2013]a)0.002b)0.003c)0.01d)0.001Correct answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
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