$${\text{Molarity}}$$   $$ = \frac{{{\text{No}}{\text{.}}\,\,{\text{of}}\,\,{\text{moles}}\,\,{\text{of}}\,\,{\text{solute}} \times {\text{1000}}}}{{{\text{Volume of solution}}\left( {{\text{in}}\,\,mL} \right)}}$$
$$\because \,\,1\,M\,{H_2}S{O_4}\,\,{\text{contains}}\,\,1\,mole\,\,{\text{of}}$$       $${{\text{H}}_2}S{O_4}\,\,{\text{in}}\,\,1000\,mL\,\,{\text{of solution}}$$
$$\therefore \,\,0.02\,M\,{H_2}S{O_4}\,\,{\text{in}}\,\,100\,\,mL\,\,{\text{of}}$$       $${\text{solution contains}}$$
$$\eqalign{ & = \frac{{0.02 \times 100}}{{1000}} \cr & = 2 \times {10^{ - 3}}mol \cr & = 2 \times {10^{ - 3}} \times 6.022 \times {10^{23}} \cr & = 12.044 \times {10^{20}}\,{\text{molecules}} \cr} $$

Given, number of molecules of urea $${\text{ = 6}}{\text{.02}} \times {\text{1}}{{\text{0}}^{20}}$$
$$\eqalign{ & \therefore \,\,{\text{Number of moles}} \cr & {\text{ = }}\frac{{6.02 \times {{10}^{20}}}}{{{N_A}}} \cr & = \frac{{6.02 \times {{10}^{20}}}}{{6.023 \times {{10}^{23}}}} \cr & = 0.999 \times {10^{ - 3}} \simeq 1 \times {10^{ - 3}}mol \cr & {\text{Volume of the solution}} \cr & = 100\,mL \cr & = \frac{{100}}{{1000}}L \cr & = 0.1\,L \cr} $$
Concentration of urea solution $$\left( {{\text{in}}\,mol\,{L^{ - 1}}} \right)$$
$$\eqalign{ & = \frac{{1 \times {{10}^{ - 3}}}}{{0.1}}mol\,{L^{ - 1}} \cr & = 1 \times {10^{ - 2}}mol\,{L^{ - 1}} \cr & = 0.01\,mol\,{L^{ - 1}} \cr} $$

$$\eqalign{ & {\text{Let }}\% {\text{ abundance of}}\,{\,^{12}}C = x \cr & \% \,\,{\text{abundance of}}\,{\,^{13}}C = 100 - x \cr & \because \,\,\frac{{12x}}{{100}} + \frac{{13\left( {100 - x} \right)}}{{100}} = 12.011 \cr & 12x + 1300 - 13x = 1201.1 \Rightarrow x = 98.9 \cr & {\text{Thus, }}\% {\text{ abundance of}}\,{\,^{12}}C\,\,{\text{is}}\,\,98.9\% . \cr} $$

Number of molecules of gas at $$STP$$
$$\eqalign{ & = \frac{{6.023 \times {{10}^{23}} \times 2.8}}{{22.4}} \cr & = 7.5 \times {10^{22}}\,{\text{molecules}} \cr} $$
Number of atoms in diatomic molecule
$$\eqalign{ & = 2 \times 7.5 \times {10^{22}} \cr & = 15 \times {10^{22}}\,{\text{atoms}} \cr} $$

In acidic medium $$MnO_4^ - $$  oxidises ferrous oxalate as follows :
$$2\,MnO_4^ - + 5{C_2}O_4^{2 - } + 16{H^ + } \to 2M{n^{2 + }} + 10C{O_2} + 8{H_2}O$$
$$\because \,5\,moles$$   of oxalate ions are oxidised by $$2\,moles$$  of $$MnO_4^ - .$$
$$\therefore \,\,1\,mole$$   of oxalate ion is oxidised by
$$\eqalign{ & = \frac{2}{5}mole\,{\text{of}}\,MnO_4^ - \cr & = 0.4\,mole\,{\text{of}}\,MnO_4^ - \cr} $$

$$Moles{\text{ of}}\,N_3^ - \,ion = \frac{{4.2}}{{42}} = 0.1$$
Each nitrogen atom has 5 valence electrons.
Therefore, total number of electrons in $$N_3^ - \,ion = 16$$
Total number of electrons in $$0.1\,mole$$   or
$$\eqalign{ & 4.2\,g\,{\text{of}}\,N_3^ - \,ion = 0.1 \times 16 \times {N_A} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1.6\,{N_A} \cr} $$

Gay Lussacs law is true only for gaseous substances.

NOTE : [ $$A{g_2}O$$  is thermally unstable and decompose on heating liberating oxygen ]
$$\eqalign{ & {\text{Mol}}{\text{.}}\,{\text{wt}}{\text{.}}\,{\text{of}}\,A{g_2}C{O_3} = 108 \times 2 + 12 + 16 \times 3 = 276g \cr & \therefore 276g\,\,{\text{of}}\,A{g_2}C{O_3}\,{\text{on}}\,{\text{heating}}\,{\text{gives}}\,{\text{residue}} \cr & = 2 \times 108 = 216g\,{\text{of}}\,Ag \cr & \therefore 2.76g\,\,{\text{of}}\,A{g_2}C{O_3}\,{\text{on}}\,{\text{heating}}\,{\text{gives}} = \frac{{216}}{{276}} \times 2.76 \cr & = 2.16g\,{\text{of}}\,Ag \cr} $$

For element $$Fe,$$  mole of atoms $$ = \frac{{69.9}}{{56}} = 1.25$$
For eIement $$O,$$  moIe of atoms $$ = \frac{{30.1}}{{16}} = 1.88$$
Mole ratio of $$Fe = \frac{{1.25}}{{1.25}} = 1,$$
MoIe ratio $$O = \frac{{1.88}}{{1.25}} = 1.5$$
Simplest whole number ratio of $$Fe$$  and $$O = 2,3$$
Empirical formula of compound $$ = F{e_2}{O_3}$$
Molecular mass of $$F{e_2}{O_3} = 160$$
$$\eqalign{ & n = \frac{{{\text{Molecular}}\,\,{\text{mass}}}}{{{\text{Empirical formula mass}}}} \cr & \,\,\,\,\, = \frac{{160}}{{160}} \cr & \,\,\,\,\, = 1 \cr} $$
Molecular formula $$ = F{e_2}{O_3}$$

$${\text{Let}}\,{\text{mass}}\,{\text{of}}\,{\text{oxygen}} = 1g,$$     $${\text{Then}}\,{\text{mass}}\,{\text{of}}\,{\text{nitrogen}} = 4g$$
$${\text{Mol}}{\text{.wt}}{\text{.of}}\,{N_2} = 28g,$$     $${\text{Mol}}{\text{.wt}}{\text{.}}\,{\text{of}}\,{O_2} = 32g$$
$$28\,g\,{\text{of}}\,{N_{2\,}}{\text{has}} = 6.02 \times {10^{23}}$$      $${\text{molecules}}\,{\text{of}}\,{\text{nitrogen}}$$
$$4g\,{\text{of}}\,{N_2}\,{\text{has}} = \frac{{6.02 \times {{10}^{23}}}}{{28}} \times 4$$       $${\text{molecules}}\,{\text{of}}\,{\text{nitrogen}}$$
$$ = \frac{{6.02 \times {{10}^{23}}}}{7}\,\,{\text{molecules}}\,{\text{of}}\,{\text{nitrogen}}$$
$$32g\,{\text{of}}\,{O_{2\,}}{\text{has}} = 6.02 \times {10^{23}}$$      $${\text{molecules}}\,{\text{of}}\,{\text{oxygen}}$$
$$\therefore 1{\text{g}}\,{\text{of}}\,{O_2}\,{\text{has}} = \frac{{6.02 \times {{10}^{23}}}}{{32}} \times 1$$       $$ = \frac{{6.02 \times {{10}^{23}}}}{{32}}\,{\text{molecules}}\,{\text{of}}\,{\text{oxygen}}$$
$${\text{Thus,}}\,{\text{ratio}}\,{\text{of}}\,{\text{molecules}}\,{\text{of}}\,{\text{oxygen}}$$       $${\text{: nitrogen}}$$
$$\eqalign{ & = \frac{{\frac{{6.02 \times {{10}^{23}}}}{{32}}}}{{\frac{{6.02 \times {{10}^{23}}}}{7}}} \cr & = 7:32 \cr} $$