Organic Chemistry Basics 2 Question 15
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15. For the estimation of nitrogen, $1.4 \mathrm{~g}$ of an organic compound was digested by Kjeldahl’s method and the evolved ammonia was absorbed in $60 \mathrm{~mL}$ of $M / 10$ sulphuric acid. The unreacted acid required $20 \mathrm{~mL}$ of $M / 10$ sodium hydroxide for complete neutralisation. The percentage of nitrogen in the compound is
======= ####15. For the estimation of nitrogen, $1.4 \mathrm{~g}$ of an organic compound was digested by Kjeldahl’s method and the evolved ammonia was absorbed in $60 \mathrm{~mL}$ of $M / 10$ sulphuric acid. The unreacted acid required $20 \mathrm{~mL}$ of $M / 10$ sodium hydroxide for complete neutralisation. The percentage of nitrogen in the compound is
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed (a) $6 %$
(b) $10 %$
(c) $3 %$
(d) $5 %$
(2014 Main)
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Answer:
Correct Answer: 15. (b)
| 17. (d) | 18. (c) | 19. (c) | 20. (b) |
|---|---|---|---|
| 21. (c) | 22. (b) | 23. (d) | 24. (a) |
| 25. (b) | 26. (d) | 27. (a) | 28. (d) |
| 29. (c) | 30. (c) | 31. (a) | 32. (a) |
| 33. (a) | 34. (a) | 35. (c) | 36. (b) |
| 37. (b) | 38. (b) | 39. (d) | 40. (b) |
| 41. (b) | 42. (d) | 43. (d) | 44. (a) |
| 45. (d) | 46. (a) | 47. (b) | 48. (b) |
| 49. (d) | 50. (c) | 51. $(\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d})$ | 52. (a) |
| 53. $(\mathrm{b}, \mathrm{c})$ | 54. $(\mathrm{b}, \mathrm{d})$ | 55. $(\mathrm{b}, \mathrm{c})$ | 56. (c,d) |
| 57. $(\mathrm{a}, \mathrm{c})$ | 58. (d) | 59. (a) | |
| 60. $\mathrm{A} \rightarrow \mathrm{r}, \mathrm{s}, \mathrm{t} ; \mathrm{B} \rightarrow \mathrm{p}, \mathrm{s} ; \mathrm{C} \rightarrow \mathrm{r}, \mathrm{s} ; \mathrm{D} \rightarrow \mathrm{q}, \mathrm{r}$ | |||
| 61. $\mathrm{A} \rightarrow \mathrm{r}, \mathrm{s} ; \mathrm{B} \rightarrow \mathrm{t} ; \mathrm{C} \rightarrow \mathrm{p}, \mathrm{q} ; \mathrm{D} \rightarrow \mathrm{r}$ | |||
| 62. hyperconjugation | |||
| 63. less | 64. cyclic | ||
| 66. triangular planar | 67. geminal, same | ||
| 68. $s p^{3}$ | 69. cylopropane | 70. propene | 71. aniline |
| 72. tert- $\mathrm{b}$ | carbonium ion, | 73. (6) | 74. (4) |
Solution:
- THINKING PROCESS This problem is based on the estimation of percentage of $\mathrm{N}$ in organic compound using Kjeldahl’s method. Use the concept of stoichiometry and follow the steps given below to solve the problem.
(a) Write the balanced chemical reaction for the conversion of $\mathrm{N}$ present in organic compound to ammonia, ammonia to ammonium sulphate and ammonium sulphate to sodium sulphate.
(b) Calculate millimoles ( $\mathrm{m}$ moles) of $\mathrm{N}$ present in organic compound followed by mass of $\mathrm{N}$ present in organic compound using the concept of stoichiometry.
(c) At last, calculate $%$ of $\mathrm{N}$ present in organic compound using formula
$$ % \text { of } \mathrm{N}=\frac{\text { Mass of } \mathrm{N} \times 100}{\text { Mass of organic compound }} $$
Mass of organic compound $=1.4 \mathrm{~g}$
Let it contain $x \mathrm{~m}$ mole of $\mathrm{N}$ atom.
$$ \begin{aligned} & \text { Organic compound } \longrightarrow \underset{3}{\mathrm{NH}{3}} \ & x \mathrm{~m} \text { mole } \ & 2 \mathrm{NH}{3}+\underset{\substack{6 \mathrm{~m} \text { mole } \ \text { initially taken }}}{\mathrm{H}{2} \mathrm{SO}{4}} \longrightarrow\left(\mathrm{NH}{4}\right){2} \mathrm{SO}{4} \ & \mathrm{H}{2} \mathrm{SO}{4}+2 \mathrm{NaOH} \longrightarrow \mathrm{Na}{2} \mathrm{SO}{4}+2 \mathrm{H}{2} \mathrm{O} \end{aligned} $$
$2 \mathrm{~m}$ mole $\mathrm{NaOH}$ reacted.
Hence, $\mathrm{m}$ moles of $\mathrm{H}{2} \mathrm{SO}{4}$ reacted in Eq. (ii) $=1$
$\Rightarrow \mathrm{m}$ moles of $\mathrm{H}{2} \mathrm{SO}{4}$ reacted from Eq. (i) $=6-1$
$$ =5 \mathrm{~m} \text { moles } $$
$\Rightarrow \mathrm{m}$ moles of $\mathrm{NH}_{3}$ in Eq. (i) $=2 \times 5=10 \mathrm{~m}$ moles
$\Rightarrow \mathrm{m}$ moles of $\mathrm{N}$ atom in the organic compound
$$ =10 \mathrm{~m} \text { moles } $$
$\Rightarrow$ Mass of $\mathrm{N}=10 \times 10^{-3} \times 14=0.14 \mathrm{~g}$
$$ \begin{aligned} % \text { of } \mathrm{N} & =\frac{\begin{array}{c} \text { Mass of } \mathrm{N} \text { present in } \ \text { organic compound } \end{array}}{\text { Mass of organic compound }} \times 100 \ \Rightarrow \quad % \text { of } \mathrm{N} & =\frac{0.14}{1.4} \times 100 \ & =10 % \end{aligned} $$
