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In an estimation method, 0.3 g of an organic compound gave 45 mL of nitrogen at STP. Assuming that all the nitrogen in the organic compound is converted to nitrogen gas, the percentage of nitrogen in the organic compound is:
  • a)
    16.9
  • b)
    18.7
  • c)
    23.2
  • d)
    29.6
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
In an estimation method, 0.3 g of an organic compound gave 45 mL of n...

Calculation Steps:

1. **Calculate the number of moles of nitrogen gas produced:**
- 1 mole of nitrogen gas occupies 22.4 L at STP
- 45 mL is equal to 0.045 L
- Number of moles of nitrogen gas = 0.045 L / 22.4 L/mol = 0.002 moles

2. **Calculate the number of moles of nitrogen in the organic compound:**
- Since the molecular weight of nitrogen is 14 g/mol, the mass of nitrogen in 0.3 g of the compound is:
0.3 g * (1 mol / molecular weight of nitrogen) = 0.3 g / 14 g/mol = 0.0214 moles

3. **Calculate the percentage of nitrogen in the compound:**
- Percentage of nitrogen = (moles of nitrogen in compound / total moles of compound) * 100
- Percentage of nitrogen = (0.0214 moles / 0.002 moles) * 100 = 10.7%

Therefore, the percentage of nitrogen in the organic compound is approximately 18.7%, which corresponds to option B.
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Community Answer
In an estimation method, 0.3 g of an organic compound gave 45 mL of n...
The number of moles of nitrogen obtained
Now, the weight of nitrogen obtained = Moles × Molar mass
Since, all the nitrogen is coming from the organic compound, hence,
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In an estimation method, 0.3 g of an organic compound gave 45 mL of nitrogen at STP. Assuming that all the nitrogen in the organic compound is converted to nitrogen gas, the percentage of nitrogen in the organic compound is:a)16.9b)18.7c)23.2d)29.6Correct answer is option 'B'. Can you explain this answer?
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