(111)

Molar mass of the compound $\left({\mathrm{C}}_4{\mathrm{N}}_2{\mathrm{H}}_{10}\right)$$=4\times 12+2\times 14+10\times 1=86{\text{g mol}}^{-1}$

$\text{Moles of compound}=\frac{\text{Mass of compound}}{\text{Molar mass of compound}}$$=\frac{0.42}{86}$ mol

$\text{Moles of N in compound}=2\times \text{Moles of compound}$$=2\times \frac{0.42}{86}$ mol

As all the N of organic compound is converted to ${\mathrm{N}}_2$ gas in Duma’s method.

Moles of ${\mathrm{N}}_2=\frac{\text{Moles of N}}{2}$
                     $=\frac{0.42}{86}\text{mol}$

Volume of $N_2$ at STP $=\text{Moles of }{N}_2\times \text{22.7\hspace{0.05em}}\mathit{\text{L}}=\frac{\mathit{0}\mathit{.}\mathit{42}}{\mathit{86}}\times \text{22.7 L}$

$=\text{0.11086\hspace{0.05em}}\mathit{\text{L}}=\text{110.8\hspace{0.05em}mL}\approx \text{111\hspace{0.05em}mL}.$

(80)

Moles of carbon in organic compound ($n_C$) = moles of carbon in ${\mathrm{CO}}_2$

$\text{=Moles of }C{O}_2=\frac{\text{Mass of }C{O}_2}{\text{Molar mass of }C{O}_2}=\frac{1.46}{44}\text{mol}$

Mass of carbon in organic compound ($W_C$)$=n_C\times \text{Molar mass of C}=\frac{1.46}{44}\times \text{12\hspace{0.05em}g}$

Percentage of carbon in organic compound

$=\frac{\text{Mass of carbon in organic compound}}{\text{Mass of organic compound}}\times 100$

$=\frac{1.46\times 12}{44\times 0.5}\times 100=79.63\%\approx 80\%$

(20)

$\text{Pressure of nitrogen gas }\left(P_{N_2}\right)=\text{Total pressure of gas}-\text{aqueous tension}$
$=(900-15)\text{mmHg}=\text{885 mmHg}=\frac{885}{760}\text{atm}$

$\text{Volume of nitrogen }\left(V_{N_2}\right)=\text{150 mL}=\text{0.15 L}$

By ideal gas equation, moles of nitrogen $\left(n_{N_2}\right)=\frac{P_{N_2}{V}_{N_2}}{RT}$

$=\frac{\frac{885}{760}\times 0.15}{0.0821\times 300}=\text{0.0071 mol}$

$\text{Mass of nitrogen }\left(W_{N_2}\right)=n_{N_2}\times \text{Molar mass of }{N}_2$

$=0.0071\times \text{28 g}=\text{0.1988 g}$

Percentage of nitrogen in organic compound $=\frac{\text{Mass of nitrogen in organic compound}}{\text{Mass of organic compound}}\times 100$

$=\frac{0.1988}{1}\times 100=19.88\%$

(18)

Pressure of nitrogen $\left(P_{N_2}\right)$ = Total pressure – aqueous tension

$=(715-15)\text{mmHg}=\frac{700}{760}\text{atm}$

Volume of nitrogen $\left(V_{N_2}\right)=\text{50\hspace{0.05em}mL}=\text{0.05\hspace{0.05em}L}$

Moles of nitrogen $\left(n_{N_2}\right)=\frac{P_{N_2}{V}_{N_2}}{RT}=\frac{\frac{700}{760}\times 0.05}{0.0821\times 300}\text{mol}$

Mass of nitrogen $\left(W_{N_2}\right)=\text{moles of nitrogen }\times \text{ molar mass of nitrogen}$

$=\frac{\frac{700}{760}\times 0.05}{0.0821\times 300}\times \text{28\hspace{0.05em}g}$

Mass of organic compound $\left(W_{o.c.}\right)=\text{292\hspace{0.05em}mg}=\text{0.292\hspace{0.05em}g}$

Percentage of nitrogen in organic compound $=\frac{W_{N_2}}{W_{o.c.}}\times 100$

$=\frac{\frac{700}{760}\times 0.05\times 28}{0.0821\times 300\times 0.292}\times 100=17.92\%\approx 18\%$