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While estimating the nitrogen present in an organic compound by Kjeldahl's method, the ammonia evolved from 0.25 g0.25 \mathrm{~g}0.25 g of the compound neutralized 2.5 mL2.5 \mathrm{~mL}2.5 mL of 2 M H2SO42 \,\mathrm{M} \,\mathrm{H}_{2} \mathrm{SO}_{4}2MH2SO4. The percentage of nitrogen present in organic compound is ______________.
NH3\mathrm{NH}_{3}NH3 gas is neutralized by 2.5 mL2.5 \mathrm{~mL}2.5 mL of 2 M H2 SO42 \mathrm{M} \,\,\mathrm{H}_{2} \mathrm{SO}_{4}2 M H2 SO4
∴\therefore∴ Moles of NH3\mathrm{NH}_{3}NH3 neutralized =2.5 × 2 × 2=2.5 \times 2 \times 2=2.5 × 2 × 2 millimole
=10 × 10−3 moles =10 \times 10^{-3} \text { moles } =10 × 10−3 moles
∴\therefore∴ Weight of N\mathrm{N}N present in the compound will be
=10 × 10−3 × 14 =0.14 g \begin{aligned} &=10 \times 10^{-3} \times 14 \\ &=0.14 \mathrm{~g} \end{aligned} =10 × 10−3 × 14 =0.14 g ∴ %\\ \therefore \% ∴ % of ' N\mathrm{N}N ' in compound
=0.140.25 × 100 =56 % \begin{aligned} &=\frac{0.14}{0.25} \times 100 \\ &=56 \% \end{aligned} =0.250.14 × 100 =56 %
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