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The weight of AgCl precipitated when a solution containing 5.85 g of NaCl is added to a solution containing 3.4 g of AgNO3 is
- a)28 g
- b)9.25 g
- c)2.870 g
- d)58 g
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
The weight of AgCl precipitated when a solution containing 5.85 g of N...
AgNO3 + NaCl → AgCl + NaNO3
No. of moles of AgNO3 = 3.4/170 = 0.02
No, of moles of NaCl = 5.85/58.5 = 0.1
Limiting reagent = AgNO3
1 mole of AgNO3 produces 1 mole of AgCl
0.02 mole of AgNO3 will produce 0.02 mole of AgCl
Weight of AgCl produced = 0.02 x 143.5 = 2.870 g.
No. of moles of AgNO3 = 3.4/170 = 0.02
No, of moles of NaCl = 5.85/58.5 = 0.1
Limiting reagent = AgNO3
1 mole of AgNO3 produces 1 mole of AgCl
0.02 mole of AgNO3 will produce 0.02 mole of AgCl
Weight of AgCl produced = 0.02 x 143.5 = 2.870 g.
Most Upvoted Answer
The weight of AgCl precipitated when a solution containing 5.85 g of N...
To find the weight of AgCl precipitated, we need to use the concept of stoichiometry and balanced chemical equations. The balanced chemical equation for the reaction between NaCl and AgNO3 is:
NaCl + AgNO3 → AgCl + NaNO3
From the equation, we can see that 1 mole of NaCl reacts with 1 mole of AgNO3 to produce 1 mole of AgCl.
1. Convert the given mass of NaCl to moles:
We know the molar mass of NaCl is 58.5 g/mol (sodium: 23 g/mol + chlorine: 35.5 g/mol).
Using the formula:
moles = mass / molar mass,
moles of NaCl = 5.85 g / 58.5 g/mol = 0.1 mol
2. Use the mole ratio from the balanced equation:
From the balanced equation, we know that 1 mole of NaCl reacts with 1 mole of AgCl. Therefore, the moles of AgCl formed will be the same as the moles of NaCl used.
3. Convert moles of AgCl to grams:
We know the molar mass of AgCl is 143.5 g/mol (silver: 107.9 g/mol + chlorine: 35.5 g/mol).
Using the formula:
mass = moles x molar mass,
mass of AgCl = 0.1 mol x 143.5 g/mol = 14.35 g
Therefore, the weight of AgCl precipitated when a solution containing 5.85 g of NaCl is added to a solution containing 3.4 g of AgNO3 is 14.35 g.
So, the correct answer is option (d) 2.870 g.
NaCl + AgNO3 → AgCl + NaNO3
From the equation, we can see that 1 mole of NaCl reacts with 1 mole of AgNO3 to produce 1 mole of AgCl.
1. Convert the given mass of NaCl to moles:
We know the molar mass of NaCl is 58.5 g/mol (sodium: 23 g/mol + chlorine: 35.5 g/mol).
Using the formula:
moles = mass / molar mass,
moles of NaCl = 5.85 g / 58.5 g/mol = 0.1 mol
2. Use the mole ratio from the balanced equation:
From the balanced equation, we know that 1 mole of NaCl reacts with 1 mole of AgCl. Therefore, the moles of AgCl formed will be the same as the moles of NaCl used.
3. Convert moles of AgCl to grams:
We know the molar mass of AgCl is 143.5 g/mol (silver: 107.9 g/mol + chlorine: 35.5 g/mol).
Using the formula:
mass = moles x molar mass,
mass of AgCl = 0.1 mol x 143.5 g/mol = 14.35 g
Therefore, the weight of AgCl precipitated when a solution containing 5.85 g of NaCl is added to a solution containing 3.4 g of AgNO3 is 14.35 g.
So, the correct answer is option (d) 2.870 g.
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The weight of AgCl precipitated when a solution containing 5.85 g of NaCl is added to a solution containing 3.4 g of AgNO3 isa)28 gb)9.25 gc)2.870 gd)58 gCorrect answer is option 'C'. Can you explain this answer?
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The weight of AgCl precipitated when a solution containing 5.85 g of NaCl is added to a solution containing 3.4 g of AgNO3 isa)28 gb)9.25 gc)2.870 gd)58 gCorrect answer is option 'C'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The weight of AgCl precipitated when a solution containing 5.85 g of NaCl is added to a solution containing 3.4 g of AgNO3 isa)28 gb)9.25 gc)2.870 gd)58 gCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The weight of AgCl precipitated when a solution containing 5.85 g of NaCl is added to a solution containing 3.4 g of AgNO3 isa)28 gb)9.25 gc)2.870 gd)58 gCorrect answer is option 'C'. Can you explain this answer?.
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