Question

0.255 g of an organic nitrogeneous compound was kjedahlised and the ammonia evolved was absorbed in 50cm³of
$\bf\dfrac{N}{10} H_2SO_4$
The excess acid required
10cm³ of one fifth Normality of
NaOH . Calculate the percentage of N in the given compound.
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Answer 1

Answer:

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Answer 2

${ \color{blue}\bigstar} \bf \:Answer$

Percentage of N in the given compound

$\rightarrow\red{16.47}$

Solution

In the present case

Mass of the organic compound ( w ) = 0.255 g

$\sf \: Volume \: of \: H_2SO_4 ( v_1) = 50 cm³$

$\sf \: Normality \: of \: H_2SO_4 (N_1) = \dfrac{1}{10}N$

$\sf Volume \: of \: NaOH \: required ( v_2) = 10 cm³$

$\sf \: Normality \: of \: NaOH (N_2) = \dfrac{1}{5}N$

$\footnotesize \sf Suppose \: V ml \: of \: H_2SO_4 \: reacts \: with \: alkali$

$: \implies \rm \dfrac{1}{10} \times v = \dfrac{1}{5} \times 10$

Therefore , we got V = 20 cm³

$\footnotesize \bf\:The \: Volume \: of \: H_2SO_4 \: reacted \: with \: NH_3 = 50 - 20 = 30 \: cm^{3}$

$\footnotesize \rm Gram \: equivalent \: of \: NH_3 \: evolved = \: Gram \: Equivalent \: of \: H_2SO_4 \: reacted \: with \: NH_3$

$: \implies \dfrac{1}{10} \times \dfrac{30}{1000}$

$\footnotesize \rm Mass \: of \: Nitrogen \: present \: in \: NH_3 \: evolved$

$\rm \therefore \: \% \: of \: N \: in \: the \: given \: compound$

$: \implies \dfrac{0.042}{0.255} \times 100 = \red{16.47}$

Conclusion

16.47 % of N is present in the Compound.

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