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An excess of AgNO3 is added to 100mL of a 0.01M solution of dichlorotetraaquachromium III chloride. The number of AgCl precipitated would be: (A)-0.002 (B)-0.003 (C)-0.01 (D)-0.001?
Most Upvoted Answer
An excess of AgNO3 is added to 100mL of a 0.01M solution of dichlorote...
**Solution:**
To find out the number of AgCl precipitated, we need to determine the reaction between AgNO3 and dichlorotetraaquachromium III chloride.
The chemical equation for the reaction can be written as follows:
AgNO3 + CrCl3.4H2O → AgCl + CrCl3.4H2O
From the balanced equation, we can see that 1 mole of AgNO3 reacts with 1 mole of CrCl3.4H2O to produce 1 mole of AgCl.
Given:
Volume of solution = 100 mL = 0.1 L
Concentration of CrCl3.4H2O = 0.01 M
**Step 1: Convert volume to moles**
Number of moles of CrCl3.4H2O = concentration × volume
= 0.01 M × 0.1 L
= 0.001 moles
**Step 2: Determine the limiting reactant**
The molar ratio between AgNO3 and CrCl3.4H2O is 1:1. Since excess AgNO3 is added, AgNO3 is not the limiting reactant. Therefore, the limiting reactant is CrCl3.4H2O.
**Step 3: Calculate the number of moles of AgCl precipitated**
From the balanced equation, we know that 1 mole of CrCl3.4H2O produces 1 mole of AgCl.
Therefore, the number of moles of AgCl precipitated = 0.001 moles
**Step 4: Calculate the mass of AgCl precipitated**
To calculate the mass of AgCl precipitated, we need to know the molar mass of AgCl.
The molar mass of AgCl = atomic mass of Ag + atomic mass of Cl
= 107.87 g/mol + 35.45 g/mol
= 143.32 g/mol
Mass of AgCl precipitated = number of moles × molar mass
= 0.001 moles × 143.32 g/mol
= 0.14332 g
**Step 5: Convert mass to volume**
To convert the mass of AgCl precipitated to volume, we need to know the density of AgCl. Assuming the density of AgCl is 5.56 g/mL, we can calculate the volume as follows:
Volume of AgCl precipitated = mass/density
= 0.14332 g/5.56 g/mL
= 0.0258 mL
Therefore, the number of AgCl precipitated is approximately 0.0258 mL.
**Answer: (D) 0.001**
To find out the number of AgCl precipitated, we need to determine the reaction between AgNO3 and dichlorotetraaquachromium III chloride.
The chemical equation for the reaction can be written as follows:
AgNO3 + CrCl3.4H2O → AgCl + CrCl3.4H2O
From the balanced equation, we can see that 1 mole of AgNO3 reacts with 1 mole of CrCl3.4H2O to produce 1 mole of AgCl.
Given:
Volume of solution = 100 mL = 0.1 L
Concentration of CrCl3.4H2O = 0.01 M
**Step 1: Convert volume to moles**
Number of moles of CrCl3.4H2O = concentration × volume
= 0.01 M × 0.1 L
= 0.001 moles
**Step 2: Determine the limiting reactant**
The molar ratio between AgNO3 and CrCl3.4H2O is 1:1. Since excess AgNO3 is added, AgNO3 is not the limiting reactant. Therefore, the limiting reactant is CrCl3.4H2O.
**Step 3: Calculate the number of moles of AgCl precipitated**
From the balanced equation, we know that 1 mole of CrCl3.4H2O produces 1 mole of AgCl.
Therefore, the number of moles of AgCl precipitated = 0.001 moles
**Step 4: Calculate the mass of AgCl precipitated**
To calculate the mass of AgCl precipitated, we need to know the molar mass of AgCl.
The molar mass of AgCl = atomic mass of Ag + atomic mass of Cl
= 107.87 g/mol + 35.45 g/mol
= 143.32 g/mol
Mass of AgCl precipitated = number of moles × molar mass
= 0.001 moles × 143.32 g/mol
= 0.14332 g
**Step 5: Convert mass to volume**
To convert the mass of AgCl precipitated to volume, we need to know the density of AgCl. Assuming the density of AgCl is 5.56 g/mL, we can calculate the volume as follows:
Volume of AgCl precipitated = mass/density
= 0.14332 g/5.56 g/mL
= 0.0258 mL
Therefore, the number of AgCl precipitated is approximately 0.0258 mL.
**Answer: (D) 0.001**
Community Answer
An excess of AgNO3 is added to 100mL of a 0.01M solution of dichlorote...
Answer is A
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An excess of AgNO3 is added to 100mL of a 0.01M solution of dichlorotetraaquachromium III chloride. The number of AgCl precipitated would be: (A)-0.002 (B)-0.003 (C)-0.01 (D)-0.001?
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An excess of AgNO3 is added to 100mL of a 0.01M solution of dichlorotetraaquachromium III chloride. The number of AgCl precipitated would be: (A)-0.002 (B)-0.003 (C)-0.01 (D)-0.001? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about An excess of AgNO3 is added to 100mL of a 0.01M solution of dichlorotetraaquachromium III chloride. The number of AgCl precipitated would be: (A)-0.002 (B)-0.003 (C)-0.01 (D)-0.001? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An excess of AgNO3 is added to 100mL of a 0.01M solution of dichlorotetraaquachromium III chloride. The number of AgCl precipitated would be: (A)-0.002 (B)-0.003 (C)-0.01 (D)-0.001?.
An excess of AgNO3 is added to 100mL of a 0.01M solution of dichlorotetraaquachromium III chloride. The number of AgCl precipitated would be: (A)-0.002 (B)-0.003 (C)-0.01 (D)-0.001? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about An excess of AgNO3 is added to 100mL of a 0.01M solution of dichlorotetraaquachromium III chloride. The number of AgCl precipitated would be: (A)-0.002 (B)-0.003 (C)-0.01 (D)-0.001? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An excess of AgNO3 is added to 100mL of a 0.01M solution of dichlorotetraaquachromium III chloride. The number of AgCl precipitated would be: (A)-0.002 (B)-0.003 (C)-0.01 (D)-0.001?.
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