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1.17 gm of sample of Na2CO3 and K2CO3 was dissolved in water to form 100ml of solution. 20 ml of this solution required 40 ml of .1M HCl for complete neutralisation .Calculate the mole % of Na2CO3 in the sample?
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1.17 gm of sample of Na2CO3 and K2CO3 was dissolved in water to form 1...
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1.17 gm of sample of Na2CO3 and K2CO3 was dissolved in water to form 1...
Given Information:
- Mass of Na2CO3 + K2CO3 = 1.17 gm
- Volume of solution = 100 ml
- Volume of HCl required for neutralization = 40 ml
- Concentration of HCl = 0.1 M
To Calculate:
Mole % of Na2CO3 in the sample
Solution:
Step 1: Calculate the number of moles of HCl used for neutralization
Molarity (M) = moles/volume (L)
Number of moles of HCl = Molarity × Volume (in L)
Number of moles of HCl = 0.1 × 0.04 = 0.004 moles
Step 2: Calculate the number of moles of Na2CO3 in the solution
From the balanced chemical equation of neutralization:
2HCl + Na2CO3 → 2NaCl + H2O + CO2
The stoichiometric ratio of HCl to Na2CO3 is 2:1
Therefore, the number of moles of Na2CO3 = 0.5 × number of moles of HCl = 0.5 × 0.004 = 0.002 moles
Step 3: Calculate the mole % of Na2CO3 in the sample
Mole % of Na2CO3 = (Number of moles of Na2CO3 / Total number of moles of Na2CO3 + K2CO3) × 100
Total number of moles of Na2CO3 + K2CO3 = mass / molar mass
Molar mass of Na2CO3 = (2 × atomic mass of Na) + atomic mass of C + (3 × atomic mass of O) = (2 × 23) + 12 + (3 × 16) = 106 g/mol
Molar mass of K2CO3 = (2 × atomic mass of K) + atomic mass of C + (3 × atomic mass of O) = (2 × 39) + 12 + (3 × 16) = 138 g/mol
Total number of moles of Na2CO3 + K2CO3 = 1.17 / (106 + 138) = 0.005 moles
Mole % of Na2CO3 = (0.002 / 0.005) × 100 = 40%
Conclusion:
The mole % of Na2CO3 in the sample is 40%.
- Mass of Na2CO3 + K2CO3 = 1.17 gm
- Volume of solution = 100 ml
- Volume of HCl required for neutralization = 40 ml
- Concentration of HCl = 0.1 M
To Calculate:
Mole % of Na2CO3 in the sample
Solution:
Step 1: Calculate the number of moles of HCl used for neutralization
Molarity (M) = moles/volume (L)
Number of moles of HCl = Molarity × Volume (in L)
Number of moles of HCl = 0.1 × 0.04 = 0.004 moles
Step 2: Calculate the number of moles of Na2CO3 in the solution
From the balanced chemical equation of neutralization:
2HCl + Na2CO3 → 2NaCl + H2O + CO2
The stoichiometric ratio of HCl to Na2CO3 is 2:1
Therefore, the number of moles of Na2CO3 = 0.5 × number of moles of HCl = 0.5 × 0.004 = 0.002 moles
Step 3: Calculate the mole % of Na2CO3 in the sample
Mole % of Na2CO3 = (Number of moles of Na2CO3 / Total number of moles of Na2CO3 + K2CO3) × 100
Total number of moles of Na2CO3 + K2CO3 = mass / molar mass
Molar mass of Na2CO3 = (2 × atomic mass of Na) + atomic mass of C + (3 × atomic mass of O) = (2 × 23) + 12 + (3 × 16) = 106 g/mol
Molar mass of K2CO3 = (2 × atomic mass of K) + atomic mass of C + (3 × atomic mass of O) = (2 × 39) + 12 + (3 × 16) = 138 g/mol
Total number of moles of Na2CO3 + K2CO3 = 1.17 / (106 + 138) = 0.005 moles
Mole % of Na2CO3 = (0.002 / 0.005) × 100 = 40%
Conclusion:
The mole % of Na2CO3 in the sample is 40%.
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1.17 gm of sample of Na2CO3 and K2CO3 was dissolved in water to form 100ml of solution. 20 ml of this solution required 40 ml of .1M HCl for complete neutralisation .Calculate the mole % of Na2CO3 in the sample?
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1.17 gm of sample of Na2CO3 and K2CO3 was dissolved in water to form 100ml of solution. 20 ml of this solution required 40 ml of .1M HCl for complete neutralisation .Calculate the mole % of Na2CO3 in the sample? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 1.17 gm of sample of Na2CO3 and K2CO3 was dissolved in water to form 100ml of solution. 20 ml of this solution required 40 ml of .1M HCl for complete neutralisation .Calculate the mole % of Na2CO3 in the sample? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1.17 gm of sample of Na2CO3 and K2CO3 was dissolved in water to form 100ml of solution. 20 ml of this solution required 40 ml of .1M HCl for complete neutralisation .Calculate the mole % of Na2CO3 in the sample?.
1.17 gm of sample of Na2CO3 and K2CO3 was dissolved in water to form 100ml of solution. 20 ml of this solution required 40 ml of .1M HCl for complete neutralisation .Calculate the mole % of Na2CO3 in the sample? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 1.17 gm of sample of Na2CO3 and K2CO3 was dissolved in water to form 100ml of solution. 20 ml of this solution required 40 ml of .1M HCl for complete neutralisation .Calculate the mole % of Na2CO3 in the sample? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1.17 gm of sample of Na2CO3 and K2CO3 was dissolved in water to form 100ml of solution. 20 ml of this solution required 40 ml of .1M HCl for complete neutralisation .Calculate the mole % of Na2CO3 in the sample?.
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