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$N{H_2} - CO - N{H_2}\xrightarrow{NaOBr} {N_2}$

Hydrazine formed from Urea is further oxidised to $N_2$ as shown below:

$N{H_2} - \mathop {\mathop C\limits^\parallel  }\limits^O  - N{H_2} \xrightarrow[\text{Hoffmann Bromamide Degradation}]{\text{NaOBr}} N{H_2} - N{H_2} \longrightarrow {N_2}$

Assuming 1 dL of blood sample contains 30 mg of Urea, calculate the volume of $N_2$ gas obtained at NTP from the sample.


From the reaction, 1 mol of Urea liberates 22.4 L $N_2$ at NTP

Or, 60 gm Urea liberates 22.4 L $N_2$ at NTP

Or, 1 gm Urea liberates $\frac {22.4}{60}$ L $N_2$ at NTP

Or, 30 mg Urea liberates $\frac{{22.4}}{{60}} \times 30 \times {10^{ - 3}}$ L $N_2$ at NTP

= $1.12 \times 10^{-2}$ L $N_2$ at NTP

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