Some Basic Concepts in Chemistry MCQ Questions & Answers in Physical Chemistry | Chemistry

TIPS/Formulae :
(i) Oxidation state of element in its free state is zero.
(ii) Sum of oxidation states of all atoms in compound is zero.
$$\eqalign{ & O.N.\,\,{\text{of}}\,\,S\,\,{\text{in}}\,\,{S_8} = 0;\,\,O.N.\,{\text{of}}\,\,S\,{\text{in}}\,{S_2}{F_2} = + 1; \cr & O.N.\,\,{\text{of}}\,\,S\,\,{\text{in}}\,{H_2}S = - 2; \cr} $$

Ammonia is formed by reduction of nitrates and nitrites with $$Zn$$  and $$NaOH.\,\, Zn$$    and caustic soda produce nascent hydrogen which reacts with nitrates to form ammonia.
\[NaN{{O}_{3}}+8{{H}^{+}}+8{{e}^{-}}\xrightarrow{Zn/NaOH}NaOH+N{{H}_{3}}+2{{H}_{2}}O\]
From the equation,
Mass of \[8\,moles\]   of electron absorbs \[85\,g\,NaN{{O}_{3}}\]
∴ Mass of $$1\,mole$$  of electron absorbs $$\frac{{85}}{8} = 10.625\,g$$    of $$NaN{O_3}.$$

$$Meq\,\,{\text{of oxalic acid in}}\,{\text{16}}{\text{.68}}\,ml$$       $$ = Meq\,\,{\text{of}}\,NaOH$$
$$\eqalign{ & = 25 \times \frac{1}{{15}} \cr & Meq\,\,{\text{of oxalic acid in }}250{\text{ }}ml \cr & = 25 \times \frac{1}{{15}} \times \frac{{250}}{{16.68}} = 24.98 \cr & \frac{{1.575}}{{\frac{{\left( {90 + 18x} \right)}}{2}}} \times 1000 = 24.98 \cr & \therefore \,\,x = 2 \cr} $$

$$\eqalign{ & {H_2}S{O_4}\,{\text{is}}\,\,{\text{dibasic}}{\text{.}} \cr & 0.1\,M\,\,{H_2}S{O_4} = 0.2N\,{H_2}S{O_4}\,\,\,\left[ {\because \,M = 2 \times N} \right] \cr & {M_{eq}}\,{\text{of}}\,{H_2}S{O_4}\,{\text{take}}{{\text{n}}}{\text{ = 100}} \times {\text{0}}{\text{.2 = 20}} \cr & {M_{eq}}\,{\text{of}}\,{H_2}S{O_4}\,\,{\text{neutralised}}\,{\text{by}}\,NaOH = 20 \times 0.5 = 10 \cr & {M_{eq}}\,{\text{of}}\,{H_2}S{O_4}\,\,{\text{neutralised}}\,{\text{by}}\,N{H_3} = 20 - 10 = 10 \cr & \% \,of\,{N_2} = \frac{{{\text{1}}{\text{.4}} \times {M_{eq}}\,{\text{of}}\,{\text{acid}}\,{\text{neutrialised}}\,{\text{by}}\,N{H_3}}}{{{\text{wt}}{\text{.}}\,\,{\text{of}}\,{\text{organic}}\,{\text{compound}}}} \cr & = \frac{{1.4 \times 10}}{{0.3}} = 46.6 \cr & \% \,{\text{of}}\,{\text{nitrogen}}\,\,{\text{in}}\,\,{\text{urea}}\, = \frac{{14 \times 2 \times 100}}{{60}} = 46.6\left[ {{\text{Mol}}{\text{.}}\,{\text{wt}}\,{\text{of}}\,{\text{urea = 60}}} \right] \cr & {\text{Similarly}}\,\% \,{\text{of}}\,{\text{Nitrogen}}\,{\text{in}}\,{\text{Benzamide}} \cr & = \frac{{14 \times 100}}{{121}} = 11.5\% \,\left[ {{C_6}{H_5}CON{H_2} = 121} \right] \cr & {\text{Acetamide}} = \frac{{14 \times 1 \times 100}}{{59}} = 23.4\% \,\left[ {C{H_3}CON{H_2} = 59} \right] \cr & {\text{Thiourea}} = \frac{{14 \times 2 \times 100}}{{76}} = 36.8\% \,\left[ {N{H_2}CSN{H_2} = 76} \right] \cr & {\text{Hence}}\,{\text{the}}\,{\text{compound}}\,{\text{must}}\,{\text{be}}\,{\text{urea}}{\text{.}} \cr} $$

$$Zn + {H_2}S{O_4} \to ZnS{O_4} + {H_2}$$
$$65\,g\,\,Zn$$   gives $$1$$ $$mole$$  of $${H_2} = 22400\,mL$$    of $${H_2}\,224\,mL$$   of $${H_2}$$ will be obtained from $$0.65\,g\,Zn.$$

The reaction that takes place is
$$NaCl + AgN{O_3} \to AgCl \downarrow + NaN{O_3}$$
$$\therefore \,\,143.5\,g$$   of $$AgCl$$  is produced from $$58.5$$  $$g$$ $$NaCl$$
$$\therefore \,\,14\,g$$   of $$AgCl$$  will be produced from $$\frac{{58.5 \times 14}}{{143.5}} = 5.70\,g\,NaCl$$
This is the amount of $$NaCl$$  in common salt; $$\% \,{\text{purity}} = \frac{{5.70}}{6} \times 100 = 95\% $$

$$\mathop {{C_6}{H_6}}\limits_{78\,g} + HN{O_3} \to \mathop {{C_6}{H_5}N{O_2}}\limits_{123\,g} + {H_2}O$$
Now since $$78g$$  of benzene on nitration give $$= 123g$$  nitrobenzene
hence $$5g$$  of benzene on nitration give
$$\frac{{123}}{{78}} \times 5 = 7.88g$$
nearest answer is (C) i.e. theoritical yield $$ = 7.88\,g$$

$$\eqalign{ & 2BC{l_3} + 3{H_2} \to 2B\, + \,6HCl \cr & or\,BC{l_3} + \frac{3}{2}{H_2} \to B + 3HCl \cr & {\text{Now,}}\,{\text{since}}\,{\text{10}}{\text{.8}}\,gm\,{\text{boron}}\,{\text{requires}}\,{\text{hydrogen}} \cr & {\text{ = }}\frac{3}{2} \times 22.4L\,at\,N.T.P \cr & {\text{hence}}\,{\text{21}}{\text{.6}}\,gm\,{\text{boron}}\,{\text{requires}}\,{\text{hydrogen}} \cr & \frac{3}{2} \times \frac{{22.4}}{{10.8}} \times 21.6 = 67.2L\,{\text{at}}\,N.T.P. \cr} $$

\[L\text{-}\underset{\left( A \right)}{\mathop{\text{amino acid}}}\,\xrightarrow[{{0}^{\circ }}C]{NaN{{O}_{2}}+HCl}{{N}_{2}}\uparrow \]
$$1\,mole\,\,{\text{of}}\,\,L{\text{ - amino acid}}$$      $$ \to {\text{1}}\,mole\,\,{\text{of}}\,\,{N_2}$$     $$ = 22400\,c{m^3}\,\,{\text{of}}\,\,{N_2}\,\,{\text{at }}STP$$
$$448\,c{m^3}\,\,{\text{of}}\,\,{N_2} \equiv 1.78\,g\,\,{\text{of}}$$      $$L{\text{ - amino acid}}$$
$$22,400\,c{m^3}\,\,{\text{of}}\,\,{N_2} \equiv x\,g\,\,{\text{of}}$$       $$L{\text{ - amino acid}}$$
$$x = \frac{{1.78 \times 22,400}}{{448}} = 89\,g\,\,{\text{of}}$$       $$L{\text{ - amino acid}}$$
$${\text{Sample of protein has}}\,\,0.25\% $$       $${\text{amino acid}}{\text{.}}$$
\[\underset{\begin{smallmatrix} 0.25 \\ 89 \end{smallmatrix}}{\mathop{\text{Amino acid}}}\,\,\,\,\,\,\,\,\,\,\,\,\underset{\begin{smallmatrix} 100 \\ y \end{smallmatrix}}{\mathop{\text{Protein}}}\,\]
$$\eqalign{ & y = \frac{{89 \times 100}}{{0.25}} \cr & \,\,\,\, = 35,600\,g/mol \cr} $$

$$21\% $$  of 1 litre is 0.21 litre.
22.4 litres = 1 $$mole$$  at $$STP$$
$$\therefore $$ 0.21 litre $${\text{ = }}\frac{{0.21}}{{22.4}} = 0.0093\,mol$$