$$IR$$  spectroscopy is used for the purification of cyclohexanone from a mixture of benzoic acid, isoamyl alcohol, cyclohexane and cyclohexanone because in $$IR$$  spectroscopy each functional group appears at a certain peak. $$IR$$  spectroscopy exploits the fact that molecules absorb specific frequencies that are characteristic of their structure.

Thus, the empirical formula of the compound is $$C{H_3}O.$$

$$\eqalign{ & V.D. = \frac{{Wt.\,{\text{of}}\,\,45\,ml.\,{\text{of}}\,\,{\text{vapours at }}NTP}}{{Wt.\,{\text{of}}\,\,45\,ml.\,{\text{of}}\,{H_2}\,\,{\text{at }}NTP}} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = \frac{{0.24\,g}}{{45ml. \times 0.000089\,g\,m{l^{ - 1}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 59.93 \cr} $$

Kjeldahl’s method is not applicable for compounds containing nitrogen in nitro and azo groups and nitrogen in ring, as $$N$$  of these compounds does not change to ammonium sulphate under these conditions.

According to combined gas equation,
$$\frac{{{p_1}{V_1}}}{{{T_1}}} = \frac{{{p_2}{V_2}}}{{{T_2}}}$$
Where, $${p_2} = $$  pressure of $${N_2}$$  at $$STP$$  $$ = 760\,mm$$
$${T_2} = $$  Temperature of $${N_2}$$  at $$STP$$  $$ = 273\,K$$
$${V_2} = ?$$
Volume of $${N_2}$$  at $$STP$$  ( By gas equation )
$$\left( {\frac{{p - {p_1}}}{{t + 273}}} \right){V_1} \times \frac{{273}}{{760}} = {V_2}$$
Where, $${p_1} = p - {p_1}$$
$$p = 715\,mm$$   ( pressure at which $${N_2}$$  collected )
$${p_1} = $$  aqueous tension of water $$ = 15\,mm$$
$${T_1} = t + 273 = 300\,K$$
$${V_1} = 55mL = $$   volume of moist nitrogen in nitrometer
$$\eqalign{ & \therefore \,\,{V_2} = \frac{{\left( {715 - 15} \right) \times 55}}{{300}} \times \frac{{273}}{{760}} \cr & = 46.098\,mL \cr} $$
$$\% $$  of nitrogen in given compound
$$\eqalign{ & = \frac{{28}}{{22400}} \times \frac{{{V_2}}}{W} \times 100 \cr & = \frac{{28}}{{22400}} \times \frac{{46.098}}{{0.35}} \times 100 \cr & = 16.45\% \cr} $$

Let the amount of the $${K_2}C{r_2}{O_7}$$   in the mixture be $$x\,g,$$  then amount of $$KMn{O_4}$$  will be $$\left( {0.5 - x} \right)g$$
$$\eqalign{ & \therefore \,\,\left( {\frac{x}{{49}} + \frac{{0.5 - x}}{{31.6}}} \right) \cr & = \frac{{100 \times 0.15}}{{1000}} \cr} $$
where $$49$$  is $$Eq.\,wt.$$  of $${K_2}C{r_2}{O_7}$$   and $$31.6$$  is $$Eq.\,wt.$$  of $$KMn{O_4}.$$
$$\eqalign{ & {\text{On solving, we get}}\,x = 0.073\,g \cr & {\text{Percentage of}}\,{K_2}C{r_2}{O_7} \cr & = \frac{{0.0732 \times 100}}{{0.5}} \cr & = 14.64\% \cr} $$

$$N{a_2}S$$  and $$NaCN,$$  formed during fusion with metallic sodium, must be removed before adding $$AgN{O_3},$$  otherwise black $$ppt.$$  due to $$N{a_2}S$$  or white precipitate due to $$AgCN$$  will be formed and thus white precipitate of $$AgCl$$  will not be identified easily.
\[\begin{align} & N{{a}_{2}}S+2AgN{{O}_{3}}\to 2NaN{{O}_{3}}+\underset{\text{Black}}{\mathop{A{{g}_{2}}S\downarrow }}\, \\ & NaCN+AgN{{O}_{3}}\to NaN{{O}_{3}}+\underset{\text{White}}{\mathop{AgCN\downarrow }}\, \\ & NaCl+AgN{{O}_{3}}\to NaN{{O}_{3}}+\underset{\text{White}}{\mathop{AgCl\downarrow }}\, \\ & N{{a}_{2}}S+2HN{{O}_{3}}\xrightarrow{\text{boil}}2NaN{{O}_{3}}+{{H}_{2}}S\uparrow \\ & NaCN+HN{{O}_{3}}\xrightarrow{\text{boil}}NaN{{O}_{3}}+HCN\uparrow \\ \end{align}\]

Mass of the substance taken $$ = 0.25\,g$$
Volume of nitrogen collected $$ = 40\,ml$$
Atmospheric pressure $$ = 725\,mm$$
Room temperature $$ = 300\,k$$
Aqueous tension at $$300k = 25\,mm$$
Actual pressure of the gas $$ = \left( {725 - 25} \right)mmHg$$
                                           $$ = 700\,mm$$
To convert the volume at experimental conditions to volume at $$STP.$$
$$\eqalign{ & {\text{Experimental value }}\,\,\,\,\,\,\,\,\,\,{\text{At }}STP \cr & {P_1} = 700\,mm\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{P_2} = 760\,mm \cr & {V_1} = 40\,ml\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{V_2} = ? \cr & {T_1} = 300\,k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{T_2} = 273\,k \cr} $$
Substituting these values in the gase eq.
$$\frac{{{P_2}{V_2}}}{{{T_2}}} = \frac{{{P_1}{V_1}}}{{{T_1}}}$$
we get, $$\frac{{760 \times {V_2}}}{{273}} = \frac{{700 \times 40}}{{300}}$$
$${V_2} = \frac{{700 \times 40}}{{300}} \times \frac{{273}}{{760}} = 33.53\,ml$$
To convert volume at $$STP$$  into mass
$$22400\,ml$$   of nitrogen at $$STP$$  weigh $$= 28g$$
∴ $$33.53\,ml$$   of nitrogen at $$STP$$  will
$${\text{weigh}} = \frac{{28 \times 33.53}}{{22400 \times 0.25}}$$
To calculate percentage of nitrogen
$$\eqalign{ & = \frac{{28 \times 33.53}}{{22400 \times 0.25}} \times 100 \cr & = 16.76\% \cr} $$