$$\eqalign{ & 1\,mole\,\,{\text{of}}\,\,C{O_2} = 44\,g \cr & 88\,g\,\,{\text{of}}\,\,C{O_2} = 2\,moles \cr & {\text{No}}{\text{. of oxygen atoms in}}\,\,2\,moles \cr & = 2 \times 2 \times {N_A} \cr & = 2 \times 2 \times 6.023 \times {10^{23}} \cr & = 24.09 \times {10^{23}} \cr} $$

Mass of $$6.023 \times {10^{23}}$$   atoms of oxygen $$ = 16\,g$$
Mass of one atom of oxygen
$$\eqalign{ & = \frac{{16}}{{6.023 \times {{10}^{23}}}} \cr & = 2.66 \times {10^{ - 23}}g \cr} $$
Mass of $${6.023 \times {{10}^{23}}}$$   atoms of nitrogen $$ = 14\,g$$
Mass of one atom of nitrogen
$$\eqalign{ & = \frac{{14}}{{6.023 \times {{10}^{23}}}} \cr & = 2.32 \times {10^{ - 23}}g \cr} $$
Mass of $$1$$ $$mole$$  of oxygen $$ = 16\,g$$
Mass of $$1 \times {10^{ - 10}}\,mole$$    of oxygen $$ = 16 \times {10^{ - 10}}$$
Mass of $$1$$ $$mole$$  of copper $$ = 63\,g$$
Mass of $$1 \times {10^{ - 10}}\,mole$$    of copper $$ = 63 \times 1 \times {10^{ - 10}}$$
                                                        $$ = 63 \times {10^{ - 10}}$$
So, the order of increasing mass is (ii) < (i) < (iii) < (iv).

\[\underset{\begin{smallmatrix} 2\times 78 \\ =156 \end{smallmatrix}}{\mathop{2{{C}_{6}}{{H}_{6}}}}\,+\underset{\begin{smallmatrix} 15\times 32 \\ =330 \end{smallmatrix}}{\mathop{15{{O}_{2}}\left( g \right)}}\,\to 12C{{O}_{2}}\left( g \right)+6{{H}_{2}}O\left( g \right)\]
$$\because \,156\,g$$   of benzene required oxygen $$ = 15 \times 22.4\,L$$
$$\therefore 1\,g$$   of benzene required oxygen $$ = \frac{{15 \times 22.4}}{{156}}L$$
$$\therefore \,\,39\,g$$   of benzene required oxygen
$$\eqalign{ & = \frac{{15 \times 22.4 \times 39}}{{156}} \cr & = 84.0\,L \cr} $$

$$\eqalign{ & ^ \circ F = \frac{9}{5}{\,^ \circ }C + 32 \cr & 200 = \frac{9}{5}\left( {^ \circ C} \right) + 32 \cr & \,\,\,\,\,\,\,\,\,\, = \left( {200 - 32} \right) \times \frac{5}{9} \cr & \,\,\,\,\,\,\,\,\,\, = {93.3^ \circ }C \cr} $$

Volume of gold foil $$ = 25 \times 40 \times 0.25\,m{m^3}$$
$$ = 250 \times {10^{ - 3}}c{m^3}$$
Mass of gold foil $$ = 19.32 \times 250 \times {10^{ - 3}}g$$
$$ = 4.83\,g$$
No. of gold atoms $$ = \frac{{4.83}}{{197}} \times {N_A}$$
$$ = 1.47 \times {10^{22}}$$

$$\eqalign{ & {\text{Molarity}} \cr & = \frac{{{W_B}\left( {{\text{in}}\,\,g} \right)}}{{{M_B}\left( {{\text{in}}\,\,g\,\,mo{l^{ - 1}}} \right) \times {\text{Volume of solution (in }}L)}} \cr & = \frac{{Conc.\left( {{\text{in}}\,\,g\,\,{L^{ - 1}}} \right)}}{{{M_B}\left( {{\text{in}}\,\,g\,\,mo{l^{ - 1}}} \right)}} \cr & = \frac{{0.9}}{{180}} \cr} $$
$$ = 0.005\,M$$     $$\left( {\because \,\,{\text{Molar mass of}}\,\,{{\text{C}}_6}{H_{12}}{O_6} = 180\,g\,mo{l^{ - 1}}} \right)$$

$$\eqalign{ & 6.023 \times {10^{23}}{\text{molecules of}}\,CO \cr & = 1\,mole\,{\text{of}}\,CO \cr & 6.02 \times {10^{24}}\,{\text{molecules of}}\,CO \cr & = 10\,moles\,{\text{of}}\,CO \cr & = 10\,g\,\,{\text{atoms}}\,{\text{of}}\,O \cr & = 5\,g\,\,{\text{molecules of }}{O_2} \cr} $$

Molar mass of sodium acetate $$\left( {C{H_3}COONa} \right) = 82\,g/mol$$
Mass of $${C{H_3}COONa}$$    required to make $$250\,mL$$   of $$0.575\,M$$   soIution $$ = \frac{{0.575 \times 82 \times 250}}{{1000}} = 11.79\,g$$

$$\eqalign{ & {\text{TIPS/Formulae :}} \cr & {\text{(i) Find change in oxidation number of }}Cr{\text{ atom}}{\text{.}} \cr & \left( {{\text{ii}}} \right){\text{Eq}}{\text{.}}\,{\text{wt}}{\text{. = }}\frac{{{\text{Molecular}}\,\,{\text{wt}}{\text{.}}}}{{{\text{change}}\,{\text{in}}\,\,O.N}} \cr & {\text{In iodometry,}}\,{K_2}C{r_2}{O_7}\,{\text{liberates }}{I_2}{\text{ from iodides }}\left( {NaI\,{\text{or}}\,KI} \right){\text{.}} \cr & {\text{Thus it is titrated with }}N{a_2}{S_2}{O_3}{\text{ solution}}{\text{.}} \cr & 2N{a_2}{S_2}{O_3} + {I_2} \to 2NaI + N{a_2}{S_4}{O_6} \cr & O.N.\,\,{\text{of}}\,\,Cr\,{\text{changes}}\,{\text{from}}\, + 6\,\left( {{\text{in}}\,{K_2}C{r_2}{O_7}} \right)\,{\text{to}}\, + 3. \cr & {\text{i}}{\text{.e}}{\text{.}}\,{\text{ + 3}}\,{\text{change}}\,{\text{for}}\,{\text{each}}\,\,Cr\,\,{\text{atom}} \cr & C{r_2}O_7^{ - - } + 14{H^ + } + 6{e^ - } \to 2C{r^{3 + }} + 7{H_2}O \cr & {\text{Thus,}}\,{\text{one}}\,{\text{mole}}\,{\text{of}}\,{K_2}C{r_2}{O_7}\,{\text{accepts}}\,\,{\text{6}}\,\,{\text{mole}}\,{\text{of}}\,{\text{electrons}}{\text{.}} \cr & \therefore {\text{Equivalent}}\,{\text{weight = }}\frac{{{\text{Molecular}}\,{\text{weight}}}}{{\text{6}}} \cr & \cr} $$

$$\eqalign{ & {\text{Weight of}}\,N{H_3} = 4.25\,g \cr & {\text{Number of moles of}} \cr & N{H_3} = \frac{{{\text{weight}}}}{{{\text{Molecular weight}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{4.25}}{{17}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 0.25\,mol \cr} $$
Number of molecules in $$0.25\,mole$$   of $$N{H_3}$$
$$ = 0.25 \times 6.023 \times {10^{23}}$$
So, number of atoms
$$\eqalign{ & = 4 \times 0.25 \times 6.023 \times {10^{23}} \cr & = 6.0 \times {10^{23}} \cr} $$