The ammonia evolved from 2 g of a compound i | vedclass

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(A) The correct option is $A$.The chemical reaction is: $2NH_3 + H_2SO_4 \longrightarrow (NH_4)_2SO_4$.Equivalents of $H_2SO_4$ used = $Molarity 	ime

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$28$

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(A) The correct option is $A$.The chemical reaction is: $2NH_3 + H_2SO_4 \longrightarrow (NH_4)_2SO_4$.Equivalents of $H_2SO_4$ used = $Molarity 	imes Volume 	imes n-factor = 2 	imes 10 	imes 10^{-3} 	imes 2 = 0.04 \ eq$.Since $1 \ mole$ of $H_2SO_4$ reacts with $2 \ moles$ of $NH_3$,the moles of $NH_3$ evolved = $2 	imes (Molarity 	imes Volume) = 2 	imes (2 	imes 10 	imes 10^{-3}) = 0.04 \ moles$.Mass of nitrogen $(N)$ in $NH_3$ = $0.04 	imes 14 \ g = 0.56 \ g$.Weight percentage of nitrogen = $\frac{	ext{Mass of } N}{	ext{Mass of compound}} 	imes 100 = \frac{0.56}{2} 	imes 100 = 28 \ \%$.

The ammonia evolved from $2 \ g$ of a compound in Kjeldahl's estimation of nitrogen neutralises $10 \ mL$ of $2 \ M \ H_2SO_4$ solution. The weight percentage of nitrogen in the compound is

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