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A mixture weighing 4.08 g of BaO and unknown carbonate XCO3 was heated strongly. The residue weighed 3.64 g. This was dissolved in 100 mL of 1 N HCl. The excess acid required 16 mL of 2.5 N NaOH solution for complete neutralization. Identify the metal M:
- a)Ag
- b)Ca
- c)Ba
- d)Na
Correct answer is option 'B'. Can you explain this answer?
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A mixture weighing 4.08 g of BaO and unknown carbonate XCO3 was heated...
From (1),(2) &(3)
m = 42
∴ metal is Ca
Most Upvoted Answer
A mixture weighing 4.08 g of BaO and unknown carbonate XCO3 was heated...
Given information:
- Mixture containing 4.08 g of BaO and unknown carbonate XCO3.
- Residue obtained after heating the mixture weighs 3.64 g.
- Residue is dissolved in 100 mL of 1 N HCl.
- Excess acid requires 16 mL of 2.5 N NaOH for complete neutralization.
To determine the metal M:
Step 1: Finding the moles of BaO
- The molecular weight of BaO (Ba = 137.33 g/mol, O = 16.00 g/mol) is 153.33 g/mol.
- Moles of BaO = mass / molecular weight = 4.08 g / 153.33 g/mol = 0.0266 mol.
Step 2: Finding the moles of CO2 produced
- BaO reacts with HCl to form BaCl2 and H2O: BaO + 2HCl → BaCl2 + H2O.
- 1 mole of BaO reacts with 2 moles of HCl to produce 1 mole of BaCl2.
- Since HCl is 1 N, it contains 1 mole of HCl in 1 L of the solution.
- Therefore, 100 mL (0.1 L) of 1 N HCl contains 0.1 moles of HCl.
- From the balanced equation, 1 mole of BaO reacts with 2 moles of HCl to produce 1 mole of BaCl2.
- So, 0.0266 moles of BaO will react with 0.0532 moles of HCl.
- This means 0.0532 moles of CO2 (molar ratio is 1:1 with HCl) will be produced.
Step 3: Finding the moles of XCO3
- The excess acid requires 16 mL (0.016 L) of 2.5 N NaOH for complete neutralization.
- From the balanced equation, 2 moles of NaOH react with 1 mole of CO2.
- Therefore, 0.016 L of 2.5 N NaOH contains 0.04 moles of NaOH.
- Since the molar ratio of NaOH to CO2 is 2:1, there must be 0.02 moles of CO2 produced.
- This means that the unknown carbonate XCO3 must have also produced 0.02 moles of CO2.
Step 4: Finding the moles of X
- Since 1 mole of XCO3 produces 1 mole of CO2, there must be 0.02 moles of XCO3.
- From the given information, the residue weighs 3.64 g.
- The molecular weight of XCO3 is equal to the sum of the atomic weights of X and CO3.
- Let's assume the atomic weight of X as M.
- Molecular weight of XCO3 = M + (12.01 + 3(16.00)) = M + 60.01 g/mol.
- Moles of XCO3 = mass / molecular weight = 3.64 g / (M + 60.01 g/mol) = 0.02 moles
- Mixture containing 4.08 g of BaO and unknown carbonate XCO3.
- Residue obtained after heating the mixture weighs 3.64 g.
- Residue is dissolved in 100 mL of 1 N HCl.
- Excess acid requires 16 mL of 2.5 N NaOH for complete neutralization.
To determine the metal M:
Step 1: Finding the moles of BaO
- The molecular weight of BaO (Ba = 137.33 g/mol, O = 16.00 g/mol) is 153.33 g/mol.
- Moles of BaO = mass / molecular weight = 4.08 g / 153.33 g/mol = 0.0266 mol.
Step 2: Finding the moles of CO2 produced
- BaO reacts with HCl to form BaCl2 and H2O: BaO + 2HCl → BaCl2 + H2O.
- 1 mole of BaO reacts with 2 moles of HCl to produce 1 mole of BaCl2.
- Since HCl is 1 N, it contains 1 mole of HCl in 1 L of the solution.
- Therefore, 100 mL (0.1 L) of 1 N HCl contains 0.1 moles of HCl.
- From the balanced equation, 1 mole of BaO reacts with 2 moles of HCl to produce 1 mole of BaCl2.
- So, 0.0266 moles of BaO will react with 0.0532 moles of HCl.
- This means 0.0532 moles of CO2 (molar ratio is 1:1 with HCl) will be produced.
Step 3: Finding the moles of XCO3
- The excess acid requires 16 mL (0.016 L) of 2.5 N NaOH for complete neutralization.
- From the balanced equation, 2 moles of NaOH react with 1 mole of CO2.
- Therefore, 0.016 L of 2.5 N NaOH contains 0.04 moles of NaOH.
- Since the molar ratio of NaOH to CO2 is 2:1, there must be 0.02 moles of CO2 produced.
- This means that the unknown carbonate XCO3 must have also produced 0.02 moles of CO2.
Step 4: Finding the moles of X
- Since 1 mole of XCO3 produces 1 mole of CO2, there must be 0.02 moles of XCO3.
- From the given information, the residue weighs 3.64 g.
- The molecular weight of XCO3 is equal to the sum of the atomic weights of X and CO3.
- Let's assume the atomic weight of X as M.
- Molecular weight of XCO3 = M + (12.01 + 3(16.00)) = M + 60.01 g/mol.
- Moles of XCO3 = mass / molecular weight = 3.64 g / (M + 60.01 g/mol) = 0.02 moles
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A mixture weighing 4.08 g of BaO and unknown carbonate XCO3 was heated strongly. The residue weighed 3.64 g. This was dissolved in 100 mL of 1 N HCl. The excess acid required 16 mL of 2.5 N NaOH solution for complete neutralization. Identify the metal M:a)Agb)Cac)Bad)NaCorrect answer is option 'B'. Can you explain this answer?
Question Description
A mixture weighing 4.08 g of BaO and unknown carbonate XCO3 was heated strongly. The residue weighed 3.64 g. This was dissolved in 100 mL of 1 N HCl. The excess acid required 16 mL of 2.5 N NaOH solution for complete neutralization. Identify the metal M:a)Agb)Cac)Bad)NaCorrect answer is option 'B'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about A mixture weighing 4.08 g of BaO and unknown carbonate XCO3 was heated strongly. The residue weighed 3.64 g. This was dissolved in 100 mL of 1 N HCl. The excess acid required 16 mL of 2.5 N NaOH solution for complete neutralization. Identify the metal M:a)Agb)Cac)Bad)NaCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A mixture weighing 4.08 g of BaO and unknown carbonate XCO3 was heated strongly. The residue weighed 3.64 g. This was dissolved in 100 mL of 1 N HCl. The excess acid required 16 mL of 2.5 N NaOH solution for complete neutralization. Identify the metal M:a)Agb)Cac)Bad)NaCorrect answer is option 'B'. Can you explain this answer?.
A mixture weighing 4.08 g of BaO and unknown carbonate XCO3 was heated strongly. The residue weighed 3.64 g. This was dissolved in 100 mL of 1 N HCl. The excess acid required 16 mL of 2.5 N NaOH solution for complete neutralization. Identify the metal M:a)Agb)Cac)Bad)NaCorrect answer is option 'B'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about A mixture weighing 4.08 g of BaO and unknown carbonate XCO3 was heated strongly. The residue weighed 3.64 g. This was dissolved in 100 mL of 1 N HCl. The excess acid required 16 mL of 2.5 N NaOH solution for complete neutralization. Identify the metal M:a)Agb)Cac)Bad)NaCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A mixture weighing 4.08 g of BaO and unknown carbonate XCO3 was heated strongly. The residue weighed 3.64 g. This was dissolved in 100 mL of 1 N HCl. The excess acid required 16 mL of 2.5 N NaOH solution for complete neutralization. Identify the metal M:a)Agb)Cac)Bad)NaCorrect answer is option 'B'. Can you explain this answer?.
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A mixture weighing 4.08 g of BaO and unknown carbonate XCO3 was heated strongly. The residue weighed 3.64 g. This was dissolved in 100 mL of 1 N HCl. The excess acid required 16 mL of 2.5 N NaOH solution for complete neutralization. Identify the metal M:a)Agb)Cac)Bad)NaCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for A mixture weighing 4.08 g of BaO and unknown carbonate XCO3 was heated strongly. The residue weighed 3.64 g. This was dissolved in 100 mL of 1 N HCl. The excess acid required 16 mL of 2.5 N NaOH solution for complete neutralization. Identify the metal M:a)Agb)Cac)Bad)NaCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of A mixture weighing 4.08 g of BaO and unknown carbonate XCO3 was heated strongly. The residue weighed 3.64 g. This was dissolved in 100 mL of 1 N HCl. The excess acid required 16 mL of 2.5 N NaOH solution for complete neutralization. Identify the metal M:a)Agb)Cac)Bad)NaCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
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