Chemistry - PRACTICAL ORGANIC CHEMISTRY Question with Solution | TestHub

Meq of ${\mathrm{H}}_2{\mathrm{SO}}_4$ used by ${\mathrm{NH}}_3=12.5\times 1\times 2=25$

$\%$ of $\mathrm{N}$ in the compound $=\frac{25\times {10}^{-3}\times 14\times 100}{0.55}=63.6$

or

Meq. of ${\mathrm{H}}_2{\mathrm{SO}}_4=$ Meq. of ${\mathrm{NH}}_3$

$12.5\times 1\times 2=25\mathrm{meq}$ of ${\mathrm{NH}}_3$

$=25$ millimoles of ${\mathrm{NH}}_3$

So Millimoles of ${}^{\text{'}}\mathrm{N}^{\text{'}}=25$

Moles of ${}^{\text{'}}\mathrm{N}^{\text{'}}=25\times {10}^{-3}$

wt. of $\mathrm{N}=14\times 25\times {10}^{-3}$

$\% \mathrm{N}=\frac{14\times 25\times {10}^{-3}}{0.55}\times 100$

$=63.66$

$\approx 64\%$