Question

0.2 g of an organic compound was analysed by Kjeldahl's method. Ammonia evolved was absorbed in 60 ml N/5 sulphuric acid. Unused acid required 40 ml of N/10 Sodium hydroxide for couple neutralisation. Find the % of Nitrogen in the compound. ​

Answer 1

answer is in attachment!

Answer 2

Answer:

The % of Nitrogen in the compound is 56%.

Explanation:

• Using  law of chemical equivalence:

According to the law of equivalence, whenever two substances react, the equivalents of one will be equal to the equivalents of other and the equivalents of any product will also be equal to that of the reactant.

Given:

Mass of compound, m= 0.2g

• According to which;
• $gram equivalents of \mathrm{NH}_{3}+ gram equivalents of \mathrm{NaOH}= gram equivalents of \mathrm{H}_{2} \mathrm{SO_{4} } \\ moles \times \mathrm{nf}+\frac{\mathrm{N}_{1} \times \mathrm{V}_{1}}{1000}=\frac{\mathrm{N}_{2} \times \mathrm{V}_{2}}{1000} \\ \mathrm{n} \times 1=\frac{8}{1000} \\\mathrm{n}=\frac{8}{1000}=8 \times 10^{-3}$
• Now,
• $1 mole of \mathrm{NH}_{3} contains 1 mole of ' \mathrm{N} '.\\ \therefore 8 \times 10^{-3} moles of \mathrm{NH}_{3} contains 8 \times 10^{-3} moles of nitrogen.$
• Hence,
• $mass of ' \mathrm{N}{ }^{\prime}= moles \times molar mass\\=8 \times 10^{-3} \times 14=0.112 \mathrm{~g}\\ \therefore \% of ' \mathrm{N}^{\prime}=\frac{0.112}{0.2} \times 100 \\=56 \%$

The % of Nitrogen in the compound is 56%.