Chemistry M.C.Q (1 Marks)

($2$) Distillation : It is used to separate volatile liquids from non-volatile impurities and the liquids having sufficient difference in their boiling point.

($3$) Chromatography : It is based on separation by using stationary and mobile phase.

($4$) Crystallization : It is based on difference in the solubilities of the compound and impurities in a suitable solvent.

$Na + C + N + S \rightarrow NaSCN$

$Fe ^{+3}+ SCN ^{\ominus} \longrightarrow[ Fe ( SCN )]^{2+}$

Experimental values, At $STP.$

$V_{1}=40 \,\mathrm{mL}$

$V_{2}=?$

$T_{1}=300 \,\mathrm{K}$

$T_{2}=273 \,\mathrm{K}$

$P_{1}^{2}=725-25=700\, \mathrm{mm}$

$P_{2}=760\, \mathrm{mm}$

$\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}$

$V_{2}=\frac{P_{1} V_{1} T_{2}}{T_{1} P_{2}}=\frac{700 \times 40 \times 273}{300 \times 760}=33.52\, \mathrm{mL}$

$22400\, \mathrm{mL}$ of $\mathrm{N}_{2}$ at $STP$ weighs $=28\, \mathrm{g}$

$\therefore 33.52\, \mathrm{mL}$ of $\mathrm{N}_{2}$ at $STP$ weighs $=\frac{28 \times 33.52}{22400}$

$=0.0419\; \mathrm{g}$

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

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

$10\, \mathrm{m} \,\mathrm{mol} \,\mathrm{H}_{2} \mathrm{SO}_{4}=20\, \mathrm{m}\, \mathrm{mol}$ of $\mathrm{NH}_{3}$

$\left[\mathrm{H}_{2} \mathrm{SO}_{4}+2 \mathrm{NH}_{3} \longrightarrow\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\right]$

$1\, \mathrm{mol}\; \mathrm{NH}_{3}$ contains $14 \,\mathrm{g}$ nitrogen $20 \times 10^{-3}\, \mathrm{mol}\; \mathrm{NH}_{3}$ contains $14 \times 20 \times 10^{-3}$ nitrogen

$0.75\, \mathrm{g}$ of sample contains

$\%$ Nitrogen $=\frac{14 \times 20 \times 10^{-3}}{0.75} \times 100=37.33 \%$

$2 \mathrm{NaCN}+\mathrm{FeSO}_{4} \longrightarrow \mathrm{Fe}(\mathrm{CN})_{2}+\mathrm{Na}_{2} \mathrm{SO}_{4}$

$F e(C N)_{2}+4 N a C N \longrightarrow N a_{4}\left[F e(C N)_{6}\right]$

$3 N a_{4}\left[F e(C N)_{6}\right]+4 F e^{3+} \longrightarrow F e_{4}\left[F e(C N)_{6}\right]_{3}+12 N a^{+}$

It is ferri-ferrocyanide

Hence, they can be separated using method of crystallization.

$B.P.$ of methanol and acetone are $338 \,K$ and $330\, K$ which have boiling points very close to each other and hence they are separated by fractional distillation method.

Collect the residue (Sand and Sulphur powder) and dissolve it in Carbon disulphide ($C{S_2}$). Sulphur dissolves in carbon disulphide but sand is insoluble in carbon disulphide. Filter the solution and evaporate it to get sulphur. Sand is dried on filter paper.

= $\frac{{35.5}}{{143.5}} \times \frac{{0.287}}{{{\rm{0}}{\rm{.099}}}} \times 100 = 71.71\% $.

colouration $[NCERT\,XI\,pg.\# \,355]$

Empirical formula = ${C_3}{H_4}$

Therefore, Empirical formula =$C{H_2}O$

= $\frac{{12 \times 0.22}}{{44 \times 0.24}} \times 100 = 25; \,C = 25, \,H = 1.66$

Total = $26.6 = 100 -26.6 = 73.4.$

Element       No. of Moles         Simple Ratio                                      

Therefore Empirical formula $ = {C_5}{H_7}N$. 

$n = \frac{{{\rm{Mol}}{\rm{. mass}}}}{{{\rm{Emp}}{\rm{. mass}}}} = \frac{{90}}{{45}} = 2$

Mol. formula = ${({C_2}{H_5}O)_2} = {C_4}{H_{10}}{O_2}$.

Element                 No. of Moles        Simple Ratio

Therefore, Empirical formula =$C{H_5}N$  or $C{H_3}N{H_2}$.

where $V$ = Volume of acid used

$N$ = Normality of acid, $W$ = Weight of substance

Hence, $CH$

Empirical formula mass of $CH = 13$

$n = \frac{{{\rm{Mol}}{\rm{. mass}}}}{{{\rm{Emp}}{\rm{.}}\,{\rm{mass}}}} = \frac{{78}}{{13}} = 6$

Molecular formula = ${(CH)_6} = {C_6}{H_6}$.

Element             No. of Moles         Simple Ratio                           

Hence, formula = ${C_2}{H_7}N$ or $(C{H_3}C{H_2}N{H_2})$ 

Empirical formula mass = $96$ Empirical formula = ${C_4}{O_3}$

$n = \frac{{290}}{{96}} = 3$

Molecular formula = ${({C_4}{O_3})_3} = {C_{12}}{O_9}$

Empirical formula =${{C}_{3}}{{H}_{7}}$ .

Elements     $\%$          No. of moles        Simple ratio

Empirical formula =${{C}_{3}}{{H}_{8}}O$ . 

Element    $\%$            No. of moles         Simple ratio

Empirical formula =${{C}_{2}}{{H}_{4}}$ . 

 Elements               No. of moles            Simple ratio

 Empirical formula = $C{{H}_{4}}O$

Elements            No. of moles       Simple ratio

Thus, Empirical formula =$C{{H}_{2}}O$ 

$ = \frac{{32}}{{233}} \times \frac{{0.35}}{{0.2595}} \times 100 = 18.52\% \,\,gm$.

In $ 60 \,gm$ of urea nitrogen present =$ 28 \,gm$

In $100 \,gm$ of urea nitrogen present  = $ 2800/60= 46.66\%$

Molecular weight of ${C_2}{H_6}O = 46$

$ = (12 \times 2 + 4 + 16) = 44$

Molecular formula $ = \frac{{{\rm{mol}}{\rm{. wt}}}}{{{\rm{eq}}{\rm{. formula wt}}{\rm{.}}}} \times \, Emp. Formula$

$ = \frac{{132.1}}{{44}} \times {\rm{Emperical \,formula}}$

$ = 3 \times {C_2}{H_4}O = {C_6}{H_{12}}{O_3}$

Empirical formula weight $ = 12 + 2 + 16 = 30$
$\therefore \,\,\,n = \frac{{{\rm{mol}}{\rm{. wt}}{\rm{.}}}}{{{\rm{empirical \,formula \,wt}}{\rm{.}}}}$

$ = \frac{{90}}{{30}} = 3$

$\therefore $ Molecular formula of the compounds

$ = {(C{H_2}O)_3}$ $ = {C_3}{H_6}{O_3}$

      $64.5$                             $28$

      $32.25$                           $28$

$64.5 \,gm$ ${C_2}{H_5}Cl$ gives $28 \,gm$ of ${C_2}{H_4}$

$32.25\,gm$ ${C_2}{H_5}Cl$ gives $ = \frac{{28 \times 32.25}}{{64.5}}$ = $14 \,gm$ of ${C_2}{H_4}$

Obtained product is $50\%$ so mass of obtained alkene $ = \frac{{14}}{2} = 7\,gm$

Their mol. wt = $4 + 16 = 20$

% wt of $C{H_4} = \frac{{{\rm{wt \,of \,}}C{H_4}}}{{{\rm{Total \,wt}}}} \times 100  = \frac{{16}}{{20}} \times 100 = 80.0\% $

$ = \frac{2}{{18}} \times \frac{{0.9}}{{0.5}} \times 100 = 20\,\% $

Since percentage of hydrogen is $20$. Therefore, remaining is carbon i.e. $80\%.$

$C{H_3}CON{H_2} + NaOHC \overset {\Delta} \longrightarrow {H_3}COONa + {H_2}O + N{H_3}$

$H = 1\,\,gm = \frac{1}{1} = 1\,\,mol$

$\therefore \,{({C_{0.87}}{H_1})_7} = {C_{6.09}}{H_7} \approx {C_6}{H_7}$

$PV = nRT$;    $PV = \frac{w}{m}RT$

$1 \times 1 = \frac{{2.4}}{m} \times 0.082 \times 400$

$m = 2.4 \times 0.082 \times 400$$ = 78.42 \approx 79$.

$ = \frac{{1.4 \times 1 \times 58}}{{1 \times 5}} = 16.2$.

$\% \mathrm{S}=\frac{32}{233} \times \frac{1.02}{2.18} \times 100=6.42 \%$

millimoles of $\mathrm{H}_{2} \mathrm{SO}_{4}=20$

milliequivalent of $\mathrm{H}_{4} \mathrm{SO}_{4}=20 \times 2=40$

Percentage of $\mathrm{N}$ in the sample $=\frac{1.4 \times 40}{2.8}=20 \%$

$\text { Ratio }=\frac{R_{f(A)}}{R_{f(C)}}=\frac{1}{2}=0.5 \text { or } 50 \times 10^{-2}$

${\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NOS}\right]^{4-} \rightarrow \text { Violet }}$

${[\mathrm{Fe}(\mathrm{SCN})]^{2+} \rightarrow \text { Blood Red }}$

$\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4 \cdot 12 \mathrm{MoO}_3 \rightarrow \text { Yellow }$

$\mathrm{Fe}^{+3} \xrightarrow{-\mathrm{SCN}} \mathrm{Fe}(\mathrm{SCN})_3($ blood red colour $)$

Appearance of blood red colour indicates presence of both nitrogen and sulphur.

Millimole of $\mathrm{H}_2 \mathrm{SO}_4 \rightarrow 10 \times 2$

So Millimole of $\mathrm{NH}_3=20 \times 2=40$

Organic $\rightarrow$  $\mathrm{NH}_3$

Compound  40 Millimole

$\therefore$ Mole of $\mathrm{N}=\frac{40}{1000}$

wt. of $\mathrm{N}=\frac{40}{1000} \times 14$

$\%$ composition of $\mathrm{N}$ in organic compound

$=\frac{40 \times 14}{1000 \times 1} \times 100$

$=56 \%$

$R _{ f }=\frac{\text { distance covered by substance from base line }}{\text { total distance covered by solvent from base line }}$

Hence, the correct order of elution in the silico gel column chromatography is $- B < C < A$

$R _{ f }=\frac{\text { Distance travelled by compound ' } X \text { ' }}{\text { Distance travelled by solvent ' } Y '}$

$=\frac{2}{8}=0.25=25 \times 10^{-2}$

$=\frac{12}{44} \times 0.792=0.216$

$\% \text { of } C \text { in compound }=\frac{0.216}{0.492} \times 100$

$=43.90 \%$

Ans:$44$

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

$60=\frac{12}{44} \times \frac{\text { mass of } CO _2 \text { formed }}{0.5} \times 100$

Mass of $CO _2$ formed $=\frac{60 \times 44 \times 0.5}{12 \times 100}\,g$

$=1.1 \text { gram }$

$=11 \times 10^{-1} \text { gram }$

mole of $Br ^{-}=$mole of $AgBr =\frac{0.376}{188}$

mass of $Br ^{-}=\frac{0.376}{188} \times 80$

$\%$ of $Br ^{-}=\frac{0.376 \times 80}{188 \times 0.4} \times 100=40 \%$

$Na + C + N \rightarrow NaCN$

$6 NaCN + FeSO _4 \rightarrow Na _4\left[ Fe ( CN )_6\right]+ Na _2 SO _4$

$Na _4\left[ Fe ( CN )_6\right]+ Fe ^{3+} \rightarrow Fe _4\left[ Fe ( CN )_6\right]_3$

(Prussian blue)

Sulphur detection by Sodium nitroprusside

$Na _2\left[ Fe ( CN )_5 NO \right]+ Na _2 S \rightarrow  Na 4\left[ Fe ( CN )_5 NOS \right]$

$\text { [Purple }]$

Phosphorus detection by ammonium molybdate

$Na _3 PO _4+3 HNO _3 \rightarrow H _3 PO _4+3 NaNO _3$

$H _3 PO _4+12\left( NH _4\right)_2 MoO _4+21 HNO _3 \rightarrow$

$\left( NH _4\right)_3 PO _4 .12 MoO _3+21 NH _4 NO _3+12 H _2 O$

$\text { (canary yellow) }$

$\text { Halogen give specific coloured ppt with }$

$AgNO _3( aq )$

$NaCl + AgNO _3( aq ) \rightarrow AgCl + NaNO _3$

$\text { (White) }$

$NaBr + AgNO _3( aq ) \rightarrow AgBr + NaNO _3$

$\text { (Pale yellow) }$

$NaI + AgNO _3( aq ) \rightarrow AgI + NaNO _3$

$\text { (Yellow) }$

NCERT $(XI)$ Vol. $2$ Page No. $359, 360. $

$C.$ Kerosene / Naphthalene $\rightarrow$ iv. Fractional distillation.

Due to different B.P. of kerosene and Naphthalene it can be separated by fractional distillation.

$D.$ $C _6 H _{12} O _6 / NaCl \rightarrow$ i. Crystallization.

$NaCl$ (ionic compound) can be crystallized.

$=\frac{32}{233} \times \frac{1.4439}{0.471} \times 100$

$=42.10$

Nearest integer $42$

Steam distillation can separate them as ortho product is steam volatile.

$(B)$ Benzoic acid + Napthalene $\rightarrow$ $(IV)$ Crystallisation

$(C)$ Water + Aniline $\rightarrow$ $(I)$ Steam distillation

$(D)$ Napthalene + Sodium chloride $\rightarrow$ $(II)$ Sublimation

Thin layer chromoatography $(TLC)$ is another type of adsorption chromatography, which involve sepration of substance of a mixture ovel a thin layer of an adsorbent coated on glass plate. A thin layer (about $0.2\,mm$ thick) of an adsorbent (silica gel) or (Alumina) in spread overa glass plate of suitable size. Hence Assertion $(A)$ is correct and Reason $(R)$ is correct explanation of $(A)$

(Bright red precipitate)

$\text { mass of } N _{2}( g ) \text { evolved }=\frac{22.4 \times 10^{-3}}{22.4} \times 28$

$=28 \times 10^{-3} \,g$

$\% \text { of } N =\frac{28 \times 10^{-3}}{0.2} \times 100=14$

$0.3g\quad\quad\quad\quad\quad\quad\quad\quad0.2g\quad\quad\quad0.1g$

$\frac{ n _{ CO _{2}}}{ n _{ H _{2} O }}=\frac{ x }{ y / 2}=\frac{0.2 / 44}{.1 / 18}$

$\frac{2 x }{ y }=\frac{36}{44}=\frac{9}{11}$

$x =\frac{9 y }{22}$

$\frac{ n _{ C _{ x } H _{ y } O_{ z } }}{ nCO _{ 2 } }=\frac{1}{ x }$

$\frac{0.3 }{12 x + y +16 z } \times \frac{44}{0.2}=\frac{1}{ x }$

$66 x =12 x + y +16 z$

$54 x = y +16 z$

$\frac{54 \times 9 y }{22}- y =16 z$

$\frac{464 y }{22}=16 z$

$z =\frac{29 y }{22}$

$C _{ x } H _{ y } O _{z}= C _{ x } H _{ y } O _{z}$

$C _{\frac{9 y}{22}} H _{ y } O _{\frac{29 y}{22}}^{22}$

$C _{9} H _{22} O _{29}$

$\,\%$ of $C =\frac{12 \times 9}{(12 \times 9+22+29 \times 16)} \times 100=\frac{108}{594} \times 100$

$18.18 \,\%$

mass of $Cl =\frac{35.5}{143.5} \times 0.4\, g$

mass $\%$ of $Cl$ in the organic compound

$=\frac{35.5 \times 0.4}{143.5 \times 0.25} \times 100$

$=39.58 \,\%$

Weight of ' $C$ ' $=\frac{0.793}{44} \times 12=0.216 \,gm$

Moles of ' $H ^{\prime}=\frac{0.442}{18} \times 2$

Weight of ' $H ^{\prime}=\frac{0.442}{18} \times 2 \times 1=0.049 \,gm$

$\therefore$ Weight of ' $O$ ' $=0.492-0.216-0.049=0.227\, gm$ $\,\%$ of ' ${ }^{\prime}{ }^{\prime}=\frac{0.227}{0.492} \times 100=46.13\, \%$

$0.5\, g \quad 0.4\, g$

$mol$ of $Br = mol$ of $AgBr =\frac{0.4}{188}$

$\%\, Br =\%\, Br =\frac{\frac{0.4}{188} \times 80}{0.5} \times 100$

$=34.04\, \%$

$\% N =\frac{1.4(2.5 \times 2 \times 2)}{0.25}=56$

Mass of $AgBr$ obtained $=0.36\,gm$

$\therefore$ Moles of $AgBr =\frac{0.36}{188}$

$\therefore$ Mass of Bromine $=\frac{0.36}{188} \times 80=0.1532\,gm$

$\therefore \% Br$ in compound $=\frac{0.1532}{0.45} \times 100=34.04 \%$

Gives $0.7938\,g\,CO _{2}=0.018$ moles

$0.4428\,g\,H _{2} O =0.0246$ moles

So moles of $C =0.018 \Rightarrow 0.216\,g$

Moles of $H =0.049 \Rightarrow 0.049\,g$

$\therefore$ wt. of Oxygen $=0.492-0.216-0.049$

$\%$ of Oxygen $=\frac{0.227}{0.492} \times 10046$ (approx.)

$P _{\text {total }}=759\,mmHg$

$P _{ N _{2}}=759-14.2=744.8\,mmHg$

$n _{ N _{2}}=\frac{744.8 \times 22.78}{760 \times 1000 \times 0.082 \times 280}=0.00097$

$W _{\text {Nitrogen }}=0.02716$

$\% N =\frac{0.02716}{0.125} \times 1000=21.728$

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

Distance travelled by the substance from

$R _{ f }=\frac{\text { reference line }( c \cdot m )}{\text { Distan ce travelled by the solvent from }}$

reference line $(c.m)$

Note : $R _{ f }$ value of different compounds are different.

on chromatogram distance travelled by cmopound is $\rightarrow 2 \,cm$

Distance travelled by solvent $=5\, cm$

So $R _{ f }=\frac{2}{5}=4 \times 10^{-1}=0.4$

So, $420 gm CO _{2} \Rightarrow \frac{12}{44} \times 420$

$\Rightarrow \frac{1260}{11} gm$ carbon

$\Rightarrow 114.545$ gram carbon

So, $\%$ of carbon $=\frac{114.545}{750} \times 100$

$=15.3 \%$

$18 gm H _{2} O \Rightarrow 2 gm H _{2}$

$210 gm \Rightarrow \frac{2}{18} \times 210$

$=23.33 gm H _{2}$

$So , \% H _{2} \Rightarrow \frac{23.33}{750} \times 100=3.11 \%$

$\approx 3 \%$

$\because 233 \,g\,\mathrm{BaSO}_{4}$ contain $\rightarrow 32\, \mathrm{~g}$ sulphur

$\therefore 1.44\, \mathrm{~g} \,\mathrm{BaSO}_{4}$ contain $\rightarrow \frac{32}{233} \times 1.44\, \mathrm{~g}$ sulphur

given : $0.471\, \mathrm{~g}$ of organic compound

$\%$ of $\mathrm{S}=\frac{32 \times 1.44}{233 \times 0.471} \times 100=41.98 \,\% \approx 42\, \%$

$\Rightarrow \mathrm{n}_{\mathrm{Br}}=\mathrm{n}_{\lambda_{\mathrm{kBt}}}=0.001\, \mathrm{~mol}$

$\Rightarrow \text { mass }_{\mathrm{Bz}}=(0.001 \times 80) \,\mathrm{gm}=0.08 \,\mathrm{gm}$

$\Rightarrow \text { mass } \%=\frac{0.08 \times 100}{0.2}=40 \,\%$

$=\left(\frac{57.5}{115}\right)=0.5\, \mathrm{~mol}$

$\Rightarrow \mathrm{C}_{7} \mathrm{H}_{17} \mathrm{~N}+\frac{45}{2} \mathrm{CuO} \rightarrow 7 \mathrm{CO}_{2}+\frac{17}{2} \mathrm{H}_{2} \mathrm{O}+\frac{1}{2} \mathrm{~N}_{2}+\frac{45}{2} \mathrm{Cu}$

$\frac{\mathrm{n}_{\mathrm{cuo}} \text { reacted }}{\left(\frac{45}{2}\right)}=\frac{\mathrm{n}_{\mathrm{C_7} \mathrm{H}_{17} \mathrm{N}} \text { reacted }}{1}$

$\Rightarrow \mathrm{n}_{\mathrm{C} \mathrm{u} O}$ reacted $=\left(\frac{45}{2}\right) \times 0.5=11.25$

$B.P.$ of aniline $=457 K$

thus can be seprated of simple distillation.

Statement $2 :$ Mixture of aniline and water seprated by simple distillation.

$\quad\quad\quad\downarrow$

$\mathrm{K} \mathrm{Fe}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$

Colloidal species

Lunar caustic $\left(\mathrm{AgNO}_{3}\right)$ is used as reagent hare to distinguish $\mathrm{Cl}^{-}, \mathrm{Br}$ and $\mathrm{I}^{-}$respectively as follows.

$\mathrm{Cl}^{-}(\mathrm{aq}) \stackrel{\mathrm{AgNO}_{3}}{\longrightarrow} \mathrm{AgCl} \downarrow_{\mathrm{ppt}}$ white

$\mathrm{Br}^{-}(\mathrm{aq})\stackrel{\mathrm{AgNO}_{3}}{\longrightarrow} \mathrm{AgBr} \downarrow_{\mathrm{ppt}} \mathrm{pale}$ yellow

$\mathrm{I}^{-}(\mathrm{aq}) \stackrel{\mathrm{AgNO}_{3}}{\longrightarrow} \mathrm{AgNO}_{3} \mathrm{AgI} \downarrow_{\mathrm{ppt}}$ Dark yellow

$0.1840\, gm$ of organic compound gave $30 \,mL$ of nitrogen which is collected at $287\, K$ And $758\,mm$ of $Hg$.

Given ; Aqueous tension at $287 \,K =14 \,mm$ of $Hg$. Hence actual pressure $=(758-14)$

$=744\, mm \text { of } Hg \text { . }$

Volume of nitrogen at $STP =\frac{273 \times 744 \times 30}{287 \times 760}$

$V =27.935 \,mL$

$\because 22400 mL$ of $N _{2}$ at STP weighs $=28\, gm .$

$\therefore 27.94 mL$ of $N _{2}$ at STP weighs $=$

$\left(\frac{28}{22400} \times 27.94\right)\, gm$

$=0.0349 \,gm$

Hence $=\left(\frac{0.0349}{0.1840} \times 100\right)$

$=18.97\, \%$

Rond off. Answer $=19 \,\%$

On balancing

$\mathrm{C}_{2} \mathrm{H}_{7} \mathrm{~N}+\frac{15}{2} \mathrm{CuO} \rightarrow 2 \mathrm{CO}_{2}+\frac{7}{2} \mathrm{H}_{2} \mathrm{O}+\frac{1}{2} \mathrm{~N}_{2}+\frac{15}{2} \mathrm{Cu}$

On comparing

$\mathrm{y}=7$

$\Rightarrow \quad$ Mass of $\mathrm{Br}=\frac{0.2397}{188} \times 80\, \mathrm{~g}$

therefore $\%$ Br in the organic compound

$=\frac{W_{\text {Br }}}{W_{T}} \times 100$

$=\frac{0.2397 \times 80}{188 \times 0.15} \times 100=0.85 \times 80$

$=68$

$\Rightarrow$ Nearest integer is $'68^{\prime}$

wt. of $\mathrm{N}=\left(\frac{42}{100} \times 0.8\right)\, \mathrm{gm}$

mole of $N=\frac{42 \times 0.8}{100 \times 14}=\frac{2.4}{100}\, \mathrm{~mol}$

moles of $\mathrm{NH}_{3}=\frac{2.4}{100}$

$2 \mathrm{NH}_{3}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}$

$\quad\downarrow\quad\quad\quad\downarrow$

$\frac{2.4}{100}$ $mole$ $\quad\frac{1.2}{100}$ $mole$

For $\mathrm{H}_{2} \mathrm{SO}_{4} \quad \frac{1.2}{100}=1 \times \mathrm{V}(\ell)$

$\Rightarrow \mathrm{V}_{\mathrm{H}_{2} \mathrm{SO}_{4}}=\frac{1.2}{100} \,\ell=12 \,\mathrm{~m} \ell$

$\Rightarrow \text { mass of chlorine }=\mathrm{n}_{\mathrm{Cl} } \times 35.5=0.0953\, \mathrm{gm}$

$\Rightarrow\, \% \text { wt of chlorine }=\frac{0.0953}{0.5} \times 100$

$\quad=19.06\, \%$

mass of organic compound $=0.172 gm$ mass of Bromine $=0.08 gm$

Hence $\%$ of Bromine $=\frac{0.08}{0.172} \times 100$

$=46.51 \%$

$\%$ of $Br =\frac{\text { wt of } AgBr }{\text { wt. of organic compound }} \times 100 \times \frac{\text { molar mass of Br }}{\text { AgBr }}$

$=\frac{1.88}{1.6} \times \frac{80}{188} \times 100=\frac{15040}{300.8}=50 \%$

Aniline $\rightarrow \mathrm{Ph}-\mathrm{NH}_{2}$, $(\mu=1.59 \mathrm{D})$

Acetophenone $\rightarrow$ Ph-O-CH $_{3}$ ( $\mu=3.05 \mathrm{D}$ )

Dipole moment : ${C}>{A}>{B}$

Hence the sequence of obtained compounds is $(\mathrm{C})\;,(\mathrm{A})$ and $(\mathrm{B})$

$H _2 O : \text { Sugar } \Rightarrow \text { Recrystallization }$

$H _2 O : \text { Aniline } \Rightarrow \text { Steam distillation }$

$H _2 O \text { : Toluene } \Rightarrow \text { Differential extraction }$

$C_{atom}= 6,\,N_{atom} = 2$

Hence $C_6H_8N_2$

$C\,\,\,\,\,\,\,\,\,\,\,x \times \frac{{25}}{M} = 2$

$H\,\,\,\,\,\,\,\,\,\,\,y \times \frac{{25}}{M} = 1$

$C_{2y}H_y  \equiv  24\,y\,gm\,C + y\,gm\,H$

                                             or

                            $24 : 1$ ratio by mass

Under the reduced pressure the liquid boil at low temperature and the temperature of decomposition will not reach. e.g. glycerol boils at $290^{\circ} \mathrm{C}$ with decomposition but at reduced pressure it boils at $180^{\circ} \mathrm{C}$ without decomposition.

$ = \frac{{wt.\,of\,Sulphur}}{{wt.\,of\,Organic\,Compound}} \times 100$

$8 = \frac{{32}}{{Wt.\,of\,compound}} \times 100$

$\therefore \,Wt.\,of\,compound = \frac{{32}}{8} \times 100 = 400\,g\,mo{l^{ - 1}}\,$

Mass of $A g B r=141 \mathrm{mg}=0.141 \mathrm{g}$

$1\; mole$ of $A g B r=1\; g$ atom of $B r$

$188 g$ of $A g B r=80 g$ of $B r$

$188 g$ of $A g B r$ contain bromine $=80 g$

$0.141 g$ of $A g B r$ contain bromine $=\frac{80}{188} \times 0.141$

This much amount of bromine present in $0.250 \mathrm{g}$ of organic compound

$\because \%$ of bromine $=\frac{80}{188} \times \frac{0.414}{0.250} \times 100=24 \%$

$(a)$ Aniline $\to $ $CN^-$

$(b)$ Benzene sulfonic acid $\to $ (figure)

$(c)$ Thiourea $\to $ $S^{2-}$

Reaction of $CN^-$ with hot and acidic solution of $FeSO_4$ lead to formation of $Fe_4[Fe(CN)_6]_3$ which is blue in colour. It contains iron in both $II$ and $III$ oxidation state.

Reaction of $S^{2-}$ with sodium nitroprusside

$N{a_2}S + N{a_2}[Fe{(CN)_5}NO] \to \mathop {N{a_4}[Fe{{(CN)}_5}NOS]}\limits_{(violet\,in\,colour)} $

Phenoxide ion on reacting with $FeCl_3$ give red colour with $FeCl_3$.

$ = 60 \times \frac{{M \times 2}}{{10}} = 12$

Mili equivalents of $NaOH = 20 \times \frac{M}{{10}} = 2$

Mili equivalents of $N{H_3} = 12 - 2 = 10$

$\% \,$ of nitrogen $ = \frac{{1.4 \times (N \times V)N{H_3}}}{{(Wt.\,of\,organic\,compound)}}$

$\frac{{1.4 \times 10}}{{1.4}} = 10$

meq. of $H_{2} S O_{4}=60 \times \frac{M}{10} \times 2=12$

meq. of $N a O H=20 \times \frac{M}{10}=2$

meq. of acid consumed $=12-2=10$

$\therefore \% \quad N=\frac{1.4 \times 10}{1.4}=10 \%$

Hence only Alanine can be used to determine percentage of nitrogen. 

$\underset{Alanine}{\mathop{\begin{matrix}
   {{H}_{2}}N-CH-COOH  \\
   |\,\,\,\,\,\,\,\,\,  \\
   \,\,\,\,\,\,\,\,\,CH{{(C{{H}_{3}})}_{2}}  \\
\end{matrix}}}\,$

$F{e^{3 + }} + SC{N^ - }\, \to \,\mathop {{{[Fe(SCN)]}^{2 + }}}\limits_{{\text{(dark red)}}} $

Element            No. of Moles        Simple Ratio                            

 Hence, $C_2H_4O$. 

The aqueous solution so obtained is boiled with concentrated nitric acid, and ammonium molybdate solution is added to it.

A yellow solution or precipitate indicates the presence of phosphorus in the organic compound.

The yellow precipitate is of ammonium phosphomolybdate

$2 \mathrm{P}+3 \mathrm{Na}_{2} \mathrm{O}_{2}+\mathrm{O}_{2} \stackrel{\text { fusion }}{\mathrm{air}} \rightarrow 2 \mathrm{Na}_{3} \mathrm{PO}_{4}$

$\mathrm{Na}_{3} \mathrm{PO}_{4}+12\left(\mathrm{NH}_{4}\right)_{3} \mathrm{MoO}_{4}+21 \mathrm{HNO}_{3}$

$\begin{aligned}\left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} \cdot 12 \mathrm{MoO}_{3}+& 21 \mathrm{NH}_{4} \mathrm{NO}_{3} \\ \text { canary yellow } &+12 \mathrm{H}_{2} \mathrm{O} \end{aligned}$

$\therefore $ $18\,gm$ water obtained from -$\frac{{1.4}}{{1.8}} \times 18 = 14$ $gm$.

Empirical formula Mass = $14$

$\therefore $ Empirical formula = $C{H_2}$.

Element           No. of Moles      Simple Ratio                                

Therefore $C{H_2}O$. 

ence, Empirical Formula = $C{H_3}$.

Fractional distillation involves repeated distillations and condensations, in a fractionating column. As a result of distillation and condensation at each point of the fractionating column, the vapours rising up become richer in more volatile component and the liquid falling back into the flask becomes richer in less volatile component. Thus, the low boiling liquid distils first while the higher boiling liquid distils afterwards.

$3NaCNS + FeC{l_3} \to \mathop {\mathop {Fe{{(CNS)}_3}}\limits_{{\rm{Ferric sulpho cyanide}}} }\limits_{{\rm{(Blood red colour)}}} + 3NaCl$

$A{g^ + }NO_3^ - \,\xrightarrow{{HCl}}Ag\,Cl\, \downarrow $  (White)

$A{g^ + }NO_3^ - \,\xrightarrow[{{H_2}S{O_4}}]{{Conc.}}N{O_2}\, \uparrow $ (Brown)

Element    $\%$      No. of Moles      Simple Ratio                            

Empirical formula = $C{H_2}N$

Empirical formula mass = $28$

Now, $ 200 \,ml$ of compound = $1 \,gm$

$22400\, ml$ of compound $\frac{1}{{200}} \times 22400 = 112$

$n = \frac{{12}}{2} = 6$

Therefore, Molecular formula $ = {(C{H_2}N)_4} = {C_4}{H_8}{N_4}$.

Element            No. of Moles           Simple Ratio