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1.20 gm of sample of Na2CO3 and K2CO3 was dissolved in water to form 100 ml of a solution. 20 ml of this solution required 40 ml of O.l N HCl for complete neutralisation. Calculate the weight of Na2CO3 in mixture. If another 20 ml of this solution is treated with excess of BaCl2 what willbe the weight of the precipitate?can you solve these?
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1.20 gm of sample of Na2CO3 and K2CO3 was dissolved in water to form 1...
**Given information:**
- Sample contains Na2CO3 and K2CO3
- Weight of sample = 1.20 gm
- Solution volume = 100 ml
- 20 ml of solution required 40 ml of 0.1 N HCl for neutralization
**Calculating the weight of Na2CO3 in the mixture:**
1. Calculate the number of moles of HCl used for neutralization:
- Volume of HCl used = 40 ml = 0.04 L
- Concentration of HCl = 0.1 N
- Number of moles of HCl = Concentration × Volume = 0.1 mol/L × 0.04 L = 0.004 mol
2. Determine the number of moles of Na2CO3:
- From the balanced chemical equation: 2 HCl + Na2CO3 → 2 NaCl + H2O + CO2
- It is clear that 2 moles of HCl react with 1 mole of Na2CO3
- Therefore, the number of moles of Na2CO3 = 0.004 mol ÷ 2 = 0.002 mol
3. Calculate the molar mass of Na2CO3:
- Molar mass of Na = 22.99 g/mol
- Molar mass of C = 12.01 g/mol
- Molar mass of O = 16.00 g/mol
- Molar mass of Na2CO3 = (2 × 22.99) + 12.01 + (3 × 16.00) = 105.99 g/mol
4. Determine the weight of Na2CO3 in the mixture:
- Weight of Na2CO3 = Number of moles × Molar mass = 0.002 mol × 105.99 g/mol = 0.212 g
Therefore, the weight of Na2CO3 in the mixture is 0.212 grams.
**Calculating the weight of the precipitate when treated with excess BaCl2:**
1. From the balanced chemical equation: BaCl2 + Na2CO3 → 2 NaCl + BaCO3
- It is clear that 1 mole of BaCl2 reacts with 1 mole of Na2CO3
2. Determine the number of moles of Na2CO3 in 20 ml of the solution:
- Volume of solution = 20 ml = 0.02 L
- Concentration of Na2CO3 = (0.212 g ÷ 1.20 g) × 1000 mg/ml = 176.67 mg/ml
- Number of moles of Na2CO3 = (Concentration × Volume) ÷ Molar mass
= (176.67 mg/ml × 0.02 L) ÷ 105.99 g/mol = 0.033 mol
3. Calculate the weight of the precipitate (BaCO3):
- From the balanced chemical equation, 1 mole of BaCO3 has a molar mass of 197.33 g/mol
- Weight of BaCO3 = Number of moles × Molar mass = 0.033 mol × 197.33 g/mol = 6.50 g
Therefore,
- Sample contains Na2CO3 and K2CO3
- Weight of sample = 1.20 gm
- Solution volume = 100 ml
- 20 ml of solution required 40 ml of 0.1 N HCl for neutralization
**Calculating the weight of Na2CO3 in the mixture:**
1. Calculate the number of moles of HCl used for neutralization:
- Volume of HCl used = 40 ml = 0.04 L
- Concentration of HCl = 0.1 N
- Number of moles of HCl = Concentration × Volume = 0.1 mol/L × 0.04 L = 0.004 mol
2. Determine the number of moles of Na2CO3:
- From the balanced chemical equation: 2 HCl + Na2CO3 → 2 NaCl + H2O + CO2
- It is clear that 2 moles of HCl react with 1 mole of Na2CO3
- Therefore, the number of moles of Na2CO3 = 0.004 mol ÷ 2 = 0.002 mol
3. Calculate the molar mass of Na2CO3:
- Molar mass of Na = 22.99 g/mol
- Molar mass of C = 12.01 g/mol
- Molar mass of O = 16.00 g/mol
- Molar mass of Na2CO3 = (2 × 22.99) + 12.01 + (3 × 16.00) = 105.99 g/mol
4. Determine the weight of Na2CO3 in the mixture:
- Weight of Na2CO3 = Number of moles × Molar mass = 0.002 mol × 105.99 g/mol = 0.212 g
Therefore, the weight of Na2CO3 in the mixture is 0.212 grams.
**Calculating the weight of the precipitate when treated with excess BaCl2:**
1. From the balanced chemical equation: BaCl2 + Na2CO3 → 2 NaCl + BaCO3
- It is clear that 1 mole of BaCl2 reacts with 1 mole of Na2CO3
2. Determine the number of moles of Na2CO3 in 20 ml of the solution:
- Volume of solution = 20 ml = 0.02 L
- Concentration of Na2CO3 = (0.212 g ÷ 1.20 g) × 1000 mg/ml = 176.67 mg/ml
- Number of moles of Na2CO3 = (Concentration × Volume) ÷ Molar mass
= (176.67 mg/ml × 0.02 L) ÷ 105.99 g/mol = 0.033 mol
3. Calculate the weight of the precipitate (BaCO3):
- From the balanced chemical equation, 1 mole of BaCO3 has a molar mass of 197.33 g/mol
- Weight of BaCO3 = Number of moles × Molar mass = 0.033 mol × 197.33 g/mol = 6.50 g
Therefore,
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1.20 gm of sample of Na2CO3 and K2CO3 was dissolved in water to form 100 ml of a solution. 20 ml of this solution required 40 ml of O.l N HCl for complete neutralisation. Calculate the weight of Na2CO3 in mixture. If another 20 ml of this solution is treated with excess of BaCl2 what willbe the weight of the precipitate?can you solve these?
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1.20 gm of sample of Na2CO3 and K2CO3 was dissolved in water to form 100 ml of a solution. 20 ml of this solution required 40 ml of O.l N HCl for complete neutralisation. Calculate the weight of Na2CO3 in mixture. If another 20 ml of this solution is treated with excess of BaCl2 what willbe the weight of the precipitate?can you solve these? 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 1.20 gm of sample of Na2CO3 and K2CO3 was dissolved in water to form 100 ml of a solution. 20 ml of this solution required 40 ml of O.l N HCl for complete neutralisation. Calculate the weight of Na2CO3 in mixture. If another 20 ml of this solution is treated with excess of BaCl2 what willbe the weight of the precipitate?can you solve these? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1.20 gm of sample of Na2CO3 and K2CO3 was dissolved in water to form 100 ml of a solution. 20 ml of this solution required 40 ml of O.l N HCl for complete neutralisation. Calculate the weight of Na2CO3 in mixture. If another 20 ml of this solution is treated with excess of BaCl2 what willbe the weight of the precipitate?can you solve these?.
1.20 gm of sample of Na2CO3 and K2CO3 was dissolved in water to form 100 ml of a solution. 20 ml of this solution required 40 ml of O.l N HCl for complete neutralisation. Calculate the weight of Na2CO3 in mixture. If another 20 ml of this solution is treated with excess of BaCl2 what willbe the weight of the precipitate?can you solve these? 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 1.20 gm of sample of Na2CO3 and K2CO3 was dissolved in water to form 100 ml of a solution. 20 ml of this solution required 40 ml of O.l N HCl for complete neutralisation. Calculate the weight of Na2CO3 in mixture. If another 20 ml of this solution is treated with excess of BaCl2 what willbe the weight of the precipitate?can you solve these? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1.20 gm of sample of Na2CO3 and K2CO3 was dissolved in water to form 100 ml of a solution. 20 ml of this solution required 40 ml of O.l N HCl for complete neutralisation. Calculate the weight of Na2CO3 in mixture. If another 20 ml of this solution is treated with excess of BaCl2 what willbe the weight of the precipitate?can you solve these?.
Solutions for 1.20 gm of sample of Na2CO3 and K2CO3 was dissolved in water to form 100 ml of a solution. 20 ml of this solution required 40 ml of O.l N HCl for complete neutralisation. Calculate the weight of Na2CO3 in mixture. If another 20 ml of this solution is treated with excess of BaCl2 what willbe the weight of the precipitate?can you solve these? in English & in Hindi are available as part of our courses for IIT JAM.
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1.20 gm of sample of Na2CO3 and K2CO3 was dissolved in water to form 100 ml of a solution. 20 ml of this solution required 40 ml of O.l N HCl for complete neutralisation. Calculate the weight of Na2CO3 in mixture. If another 20 ml of this solution is treated with excess of BaCl2 what willbe the weight of the precipitate?can you solve these?, a detailed solution for 1.20 gm of sample of Na2CO3 and K2CO3 was dissolved in water to form 100 ml of a solution. 20 ml of this solution required 40 ml of O.l N HCl for complete neutralisation. Calculate the weight of Na2CO3 in mixture. If another 20 ml of this solution is treated with excess of BaCl2 what willbe the weight of the precipitate?can you solve these? has been provided alongside types of 1.20 gm of sample of Na2CO3 and K2CO3 was dissolved in water to form 100 ml of a solution. 20 ml of this solution required 40 ml of O.l N HCl for complete neutralisation. Calculate the weight of Na2CO3 in mixture. If another 20 ml of this solution is treated with excess of BaCl2 what willbe the weight of the precipitate?can you solve these? theory, EduRev gives you an
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