(2)

$\%\text{ of C}=\frac{12}{44}\times \frac{{\text{Mass of CO}}_2 \text{formed}}{\text{Mass of compound taken}}\times 100$

              $=\frac{12\times 0.2\times 100}{44\times 0.3}=18.18\%$

$\%\text{ of H}=\frac{2}{18}\times \frac{{\text{Mass of H}}_2\text{O formed}}{\text{Mass of compound taken}}\times 100$

              $=\frac{2\times 0.1\times 100}{18\times 0.3}=3.70\%$

(3)

Kjeldahl’s method is not applicable to compounds containing nitrogen in nitro group, azo groups and nitrogen present in the ring (e.g., pyridine).

(1)

Mass of organic compound = 0.25 g  

Experimental values, At STP  

$V_1=40\mathrm{mL}, V_2=?, T_1=300\mathrm{K}, T_2=273\mathrm{K}$

$P_1=725-25=700\mathrm{mm}, P_2=760\mathrm{mm}$

$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2} \Rightarrow V_2=\frac{P_1{V}_1{T}_2}{T_1{P}_2}=\frac{700\times 40\times 273}{300\times 760}=33.52$ $\mathrm{mL}$

22400 mL of ${\mathrm{N}}_2$ at STP weighs = 28 g  

$\therefore$  $33.52 \mathrm{mL} {\text{of N}}_2 \text{at STP weighs}=\frac{28\times 33.52}{22400}=0.0419$ $\mathrm{g}$

$\% \text{of N}=\frac{\text{Mass of nitrogen at STP}}{\text{Mass of organic compound taken}}\times 100$

              $=\frac{0.0419}{0.25}\times 100=16.76\%$

(1)

${\mathrm{H}}_2{\mathrm{SO}}_4+2{\mathrm{NH}}_3⟶{\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4$

10 mL of 1 M ${\mathrm{H}}_2{\mathrm{SO}}_4$ = 10 mmol                                    [$\because M\times V_{\left(\mathrm{mL}\right)}=\mathrm{mmol}$]  

Acid used for the absorption of ammonia $=10 \mathrm{mL} \text{of} 2N \left(\text{or 1 M}\right) {\mathrm{H}}_2{\mathrm{SO}}_4$

$\% \text{of N}=\frac{1.4\times N\times V}{W}=\frac{1.4\times 2\times 10}{0.75}=37.33\%$