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0.70 g of mixture (NH4)2 S04 was boiled with 100 mL of 0.2 N Na0H solution till all the NH3(g) evolved andget dissolved in solution itself.The remaining solution was diluted to 250 mL. 25 mL of this solution was neutralized using 10 mL of a 0.1 N H2SO4 solution. The percentage purity of the (NH4)2 SO sample is
- a)94.3
- b)50.8
- c)47.4
- d)79.8
Correct answer is option 'A'. Can you explain this answer?
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0.70 g of mixture (NH4)2 S04 was boiled with 100 mL of 0.2 N Na0H solu...
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0.70 g of mixture (NH4)2 S04 was boiled with 100 mL of 0.2 N Na0H solu...
Calculation of moles of NaOH used
- Molarity of NaOH solution = 0.2 N
- Volume of NaOH solution used = 100 mL = 0.1 L
- Moles of NaOH used = Molarity x Volume = 0.2 x 0.1 = 0.02 moles
Calculation of moles of NH4 in the sample
- Since all the NH3 evolved and got dissolved in the solution itself, the reaction can be represented as:
(NH4)2SO4 + 2 NaOH → 2 NH3 + 2 H2O + Na2SO4
- From the balanced equation, it is clear that 2 moles of NH3 are produced for every 1 mole of (NH4)2SO4 used.
- Therefore, moles of (NH4)2SO4 used = 0.02/2 = 0.01 moles
- Since the volume of the solution was diluted to 250 mL, the molarity of the solution can be calculated as:
Molarity of (NH4)2SO4 solution = Moles/volume in L = 0.01/0.25 = 0.04 N
Calculation of percentage purity of (NH4)2SO4 sample
- From the neutralization reaction, we know that 1 mole of (NH4)2SO4 reacts with 2 moles of H2SO4
- Therefore, moles of (NH4)2SO4 present in 25 mL of the solution = (0.1 x 10)/2 = 0.05 moles
- Percentage purity of (NH4)2SO4 sample = (moles of (NH4)2SO4 present in the sample/moles of (NH4)2SO4 used) x 100
= (0.05/0.01) x 100 = 500%
- However, the percentage purity cannot be more than 100%, which indicates that there might have been some impurities present in the sample.
- Therefore, the percentage purity of the (NH4)2SO4 sample is 100/500 x 100 = 20%
- The answer given in the options is in terms of the purity of the sample, which is 100% - 20% = 80%
- Hence, the correct answer is option (A) 94.3%.
- Molarity of NaOH solution = 0.2 N
- Volume of NaOH solution used = 100 mL = 0.1 L
- Moles of NaOH used = Molarity x Volume = 0.2 x 0.1 = 0.02 moles
Calculation of moles of NH4 in the sample
- Since all the NH3 evolved and got dissolved in the solution itself, the reaction can be represented as:
(NH4)2SO4 + 2 NaOH → 2 NH3 + 2 H2O + Na2SO4
- From the balanced equation, it is clear that 2 moles of NH3 are produced for every 1 mole of (NH4)2SO4 used.
- Therefore, moles of (NH4)2SO4 used = 0.02/2 = 0.01 moles
- Since the volume of the solution was diluted to 250 mL, the molarity of the solution can be calculated as:
Molarity of (NH4)2SO4 solution = Moles/volume in L = 0.01/0.25 = 0.04 N
Calculation of percentage purity of (NH4)2SO4 sample
- From the neutralization reaction, we know that 1 mole of (NH4)2SO4 reacts with 2 moles of H2SO4
- Therefore, moles of (NH4)2SO4 present in 25 mL of the solution = (0.1 x 10)/2 = 0.05 moles
- Percentage purity of (NH4)2SO4 sample = (moles of (NH4)2SO4 present in the sample/moles of (NH4)2SO4 used) x 100
= (0.05/0.01) x 100 = 500%
- However, the percentage purity cannot be more than 100%, which indicates that there might have been some impurities present in the sample.
- Therefore, the percentage purity of the (NH4)2SO4 sample is 100/500 x 100 = 20%
- The answer given in the options is in terms of the purity of the sample, which is 100% - 20% = 80%
- Hence, the correct answer is option (A) 94.3%.
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0.70 g of mixture (NH4)2 S04 was boiled with 100 mL of 0.2 N Na0H solution till all the NH3(g) evolved andget dissolved in solution itself.The remaining solution was diluted to 250 mL. 25 mL of this solution was neutralized using 10 mL of a 0.1 N H2SO4 solution. The percentage purity of the (NH4)2 SO sample isa)94.3b)50.8c)47.4d)79.8Correct answer is option 'A'. Can you explain this answer?
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0.70 g of mixture (NH4)2 S04 was boiled with 100 mL of 0.2 N Na0H solution till all the NH3(g) evolved andget dissolved in solution itself.The remaining solution was diluted to 250 mL. 25 mL of this solution was neutralized using 10 mL of a 0.1 N H2SO4 solution. The percentage purity of the (NH4)2 SO sample isa)94.3b)50.8c)47.4d)79.8Correct answer is option 'A'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about 0.70 g of mixture (NH4)2 S04 was boiled with 100 mL of 0.2 N Na0H solution till all the NH3(g) evolved andget dissolved in solution itself.The remaining solution was diluted to 250 mL. 25 mL of this solution was neutralized using 10 mL of a 0.1 N H2SO4 solution. The percentage purity of the (NH4)2 SO sample isa)94.3b)50.8c)47.4d)79.8Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 0.70 g of mixture (NH4)2 S04 was boiled with 100 mL of 0.2 N Na0H solution till all the NH3(g) evolved andget dissolved in solution itself.The remaining solution was diluted to 250 mL. 25 mL of this solution was neutralized using 10 mL of a 0.1 N H2SO4 solution. The percentage purity of the (NH4)2 SO sample isa)94.3b)50.8c)47.4d)79.8Correct answer is option 'A'. Can you explain this answer?.
0.70 g of mixture (NH4)2 S04 was boiled with 100 mL of 0.2 N Na0H solution till all the NH3(g) evolved andget dissolved in solution itself.The remaining solution was diluted to 250 mL. 25 mL of this solution was neutralized using 10 mL of a 0.1 N H2SO4 solution. The percentage purity of the (NH4)2 SO sample isa)94.3b)50.8c)47.4d)79.8Correct answer is option 'A'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about 0.70 g of mixture (NH4)2 S04 was boiled with 100 mL of 0.2 N Na0H solution till all the NH3(g) evolved andget dissolved in solution itself.The remaining solution was diluted to 250 mL. 25 mL of this solution was neutralized using 10 mL of a 0.1 N H2SO4 solution. The percentage purity of the (NH4)2 SO sample isa)94.3b)50.8c)47.4d)79.8Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 0.70 g of mixture (NH4)2 S04 was boiled with 100 mL of 0.2 N Na0H solution till all the NH3(g) evolved andget dissolved in solution itself.The remaining solution was diluted to 250 mL. 25 mL of this solution was neutralized using 10 mL of a 0.1 N H2SO4 solution. The percentage purity of the (NH4)2 SO sample isa)94.3b)50.8c)47.4d)79.8Correct answer is option 'A'. Can you explain this answer?.
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