Final answer:
By using Kjeldahl's method, the percentage of nitrogen in the organic compound is calculated to be 19.07%, which does not match the given answer choices, suggesting a possible typo or error.
Step-by-step explanation:
The student is asking to calculate the percentage of nitrogen in an organic compound by using Kjeldahl's method. Initially, 0.6g of the compound produced ammonia, which required 90 ml of N/9 H2SO4 for absorption. Since there was excess acid, it took 20 ml of N/10 NaOH to neutralize it. To find the amount of acid neutralized by the ammonia, we subtract the amount of NaOH used from the amount of H2SO4 added:
- 90 ml of N/9 H2SO4 = 10 ml of N H2SO4
- 20 ml of N/10 NaOH = 2 ml of N NaOH
- So, the ammonia has neutralized 10 ml - 2 ml = 8 ml of N H2SO4.
Since the reaction between H2SO4 and NH3 is 1:2, the moles of NH3 are double the moles of H2SO4 neutralized. Therefore:
- 8 ml of N H2SO4 = 8/1000 moles = 8 x 10^-3 moles of H2SO4
- Which means 16 x 10^-3 moles of NH3
- Since the molar mass of NH3 is 17 g/mol, the mass of NH3 from the 0.6 g sample is 16 x 10^-3 moles x 17 g/mol = 0.272 g NH3, which contains 82 x 10^-3 moles x 14 g/mol = 0.1144 g of N
Lastly, to find the percentage of nitrogen:
- (0.1144 g / 0.6 g) x 100 = 19.07%
However, none of the given answer choices match this correct result, indicating that there may have been a typo in the question options or a misinterpretation during the process.