Question

1.4g of an organic compound was digested according to Kjeldahl’s method and the ammonia evolved was absorbed in 60 mL of M/10${{H}_{2}}S{{O}_{4}}$ solution. The excess sulphuric acid required 20 mL of M/10 $NaOH$ solution for neutralization. The percentage of nitrogen in the compound is:
A) 3
B) 5
C) 24
D) 10

Answer 1

Hint: The answer here is based on the Kjeldah’s method which has the formula for calculating percentage of the substance in the compound which is given by $\%=\dfrac{1.4\times V\times N}{W}$

Complete step by step solution:
We are familiar with the Kjeldahl method concept in the organic chemistry classes and have come across the digestion method.
Now we shall see what the method is and then find the percentage of substance formed.
- Kjeldahl’s method consists of heating the sample at 360$^{0}C$ to 410 with the concentrated sulphuric acid which decomposes the organic sample by oxidation reaction that liberates the reduced nitrogen in the form of ammonium sulphate.
The formula to calculate the amount of nitrogen formed can be calculated using the formula,
$%N=\dfrac{1.4\times V\times N}{W}$
where, V is the volume and N is the normality which together now as gram equivalent weight of ammonia
W is the weight of the organic compound where W = 1.4g
Now we have to find the gram equivalent weight and for this let us see the reaction that is taking place when ammonia is reacted with sulphuric acid which is shown below,
\[N{{H}_{3}}+{{H}_{2}}S{{O}_{4}}\to {{(N{{H}_{4}})}_{2}}S{{O}_{4}}\]
To get the gram equivalent weight, we have to multiply by 2 for gram equivalent calculation as two hydrogens are produced to convert one mole of ammonia.
Now since 60 mL of ammonia was absorbed in 1/10 M, ${{H}_{2}}S{{O}_{4}}$, we have
Gram equivalent weight = $60\times \dfrac{1}{10}=6\times 2=12g$
Now, to react the unreacted acid 20 mL of 1/10 M $NaOH$ was used and therefore the reaction can be written as,
\[{{H}_{2}}S{{O}_{4}}+NaOH\to N{{a}_{2}}S{{O}_{4}}+{{H}_{2}}O\]
Now, the gram equivalent for this is$=20\times \dfrac{1}{10}=2g$
Since, out of 12g, 2g of the acid was unreacted, we have the total gram equivalent weight of ammonia as, $12 -2 = 10g$
Thus, NV = 10g.
Substituting the values in the actual equation we have,
$%N=\dfrac{1.4\times 10}{1.4}$
$\%N=10\%$
Therefore, the amount of nitrogen present in the compound is$10\%$

Therefore, the correct answer is option D) 10.

Note: Kjeldahl’s method is also called by the name Kjeldahl’s digestion which is the method used to determine amount of nitrogen which is contained in the organic substances in addition to the nitrogen contained in the inorganic compounds that is ammonia and ammonium ion $(N{{H}_{3}}/N{{H}_{4}}^{+})$.