$$\eqalign{ & {\text{Given, molarity of solution = 2}} \cr & {\text{Volume of solution}} \cr & = 250\,mL \cr & = \frac{{250}}{{1000}} \cr & = \frac{1}{4}L \cr & {\text{Molar mass of}} \cr & HN{O_3} = 1 + 14 + 3 \times 16 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 63\,g\,mo{l^{ - 1}} \cr & \because \,{\text{Molarity}} \cr & = \frac{{{\text{Weight of}}\,HN{O_3}}}{{{\text{Molecular mass of}}\,HN{O_3} \times {\text{volume of solution}}\,\left( L \right)}} \cr} $$

$$\therefore \,\,{\text{Weight of}}\,HN{O_3}$$     $$ = {\text{molarity}} \times {\text{molecular mass}} \times {\text{volume }}(L)$$
$$\eqalign{ & = 2 \times 63 \times \frac{1}{4}g \cr & = 31.5\,g \cr & {\text{It is the weight of }}100\% {\text{ }}HN{O_3} \cr & {\text{But the given acid is }}70\% {\text{ }}HN{O_3} \cr & \therefore \,\,{\text{Its weight}}\,{\text{ = }}\,{\text{31}}{\text{.5}} \times \frac{{100}}{{70}}\,g \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 45\,g \cr} $$

$${\text{Molar mass of}}\,\,{O_2} = 32\,\,g\,mo{l^{ - 1}}$$
$$32\,\,g\,\,{\text{of}}\,\,{O_2}$$   $$ = 6.023 \times {10^{23}}\,\,{\text{molecules}}$$
$$\eqalign{ & 40\,\,g\,\,{\text{of}}\,\,{O_2} = \frac{{6.023 \times {{10}^{23}} \times 40}}{{32}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 7.529 \times {10^{23}}{\text{molecules}} \cr} $$
$${\text{Mass of}}\,\,6.023 \times {10^{23}}$$     $${\text{molecules}}\,\,{\text{of}}\,\,C{O_2} = 44\,g$$
$${\text{Mass of}}\,\,7.529 \times {10^{23}}$$     $${\text{molecules}}\,\,{\text{of}}\,\,C{O_2}$$    $$ = \frac{{44 \times 7.529 \times {{10}^{23}}}}{{6.023 \times {{10}^{23}}}} = 55\,g$$

$$\eqalign{ & {\text{Average of atomic weight}} \cr & \frac{{\% \,{\text{of}}{\,^{10}}B \times atomic{\text{ }}mass{\text{ of}}{\,^{10}}B + \% \,{\text{of}}{\,^{11}}B \times \,\,atomic{\text{ }}mass{\text{ of}}{\,^{11}}B\,}}{{\% \,{\text{of}}{\,^{10}}B + \% \,{\text{of}}{\,^{11}}B}} \cr & = \frac{{19 \times 10 + 81 \times 11}}{{19 + 81}} \cr & = \frac{{190 + 891}}{{100}} \cr & = 10.81 \cr} $$

$${\text{Density}} = \frac{{{\text{Mass}}}}{{{\text{Volume}}}}$$
$${\text{For water}}:1 = \frac{{18}}{{{V_{{H_2}O}}}}$$     $$\left[ {\because \,\,{\text{Molar mass of}}\,\,{{\text{H}}_2}O = 18\,g\,mo{l^{ - 1}}} \right]$$
$${V_{{H_2}O}} = 18\,L \Rightarrow {N_A}\,\,{\text{molecules}}$$
$${\text{For propanol}}\,\,{\text{:}}\,\,0.925 = \frac{{60}}{{{V_{{\text{propanol}}}}}}$$       $$\left[ {\because \,\,{\text{Molar mass of}}\,\,C{H_3}C{H_2}C{H_2}OH = 60\,g\,mo{l^{ - 1}}} \right]$$
$$\eqalign{ & {V_{{\text{propanol}}}} = 64.86\,L \Rightarrow {N_A}\,{\text{molecules}} \cr & {\text{Ratio}}:\frac{{{V_{{\text{propanol}}}}}}{{{V_{{H_2}O}}}} = \frac{{64.86}}{{18}} \cr & {V_{{\text{propanol}}}} = 3.603 \times {V_{{H_2}O}} \cr} $$
$$ \Rightarrow {V_{{\text{propanol}}}} = 3.603 \times 210$$      $$ = 756.63 \simeq 757\,mL$$

$$\eqalign{ & \because \,\,18g,\,{H_2}O\,{\text{contains}} = 2gm\,H \cr & \therefore \,\,0.72\,gm\,{H_2}O\,{\text{contains}} = \frac{2}{{18}} \times 0.72gm = 0.08\,gm\,H \cr & \because \,\,44\,gm\,C{O_2}\,{\text{contains}} = 12\,gm\,C \cr & \therefore 3.08\,gm\,C{O_2}\,{\text{contains}} = \frac{{12}}{{44}} \times 3.08 = 0.84\,gm\,C \cr & \therefore C:H = \frac{{0.84}}{{12}}:\frac{{0.08}}{1} = 0.07:0.08 = 7:8 \cr & \therefore \,\,{\text{Empirical formula}} = {C_7}{H_8} \cr} $$

TIPS/Formulae :
$${\text{Molality}} = \frac{{{\text{Number}}\,{\text{of}}\,{\text{moles}}\,{\text{of}}\,{\text{solute}}}}{{{\text{Mass}}\,{\text{of}}\,{\text{solvent}}\,{\text{in}}\,kg}}$$
A molal solution is one which contains one mole of
solute per 1000 $$g$$ of solvent. $$\left\{ {\because {\text{lm}} = \frac{{{\text{1mole}}}}{{{\text{1}}kg}}} \right\}$$

Volume of oxygen in $$1$$ $$L$$ of air
$$\eqalign{ & {\text{ = }}\frac{{21}}{{100}} \times 1000 \cr & = 210\,mL \cr} $$
$$\because \,\,22400{\text{ }}mL$$    volume at $$STP$$  is occupied by oxygen
$${\text{ = }}1\,mole$$
Therefore, number of moles occupied by $$210$$  $$mL$$
$$\eqalign{ & = \frac{{210}}{{22400}} \cr & = 0.0093\,mol \cr} $$

Let weight of $$C$$  be $$x\,g,$$  then $$S$$  will be $$\left( {14 - x} \right)g$$
$$\eqalign{ & \frac{{\frac{x}{{12}}}}{{\frac{{\left( {14 - x} \right)}}{{32}}}} = \frac{2}{1} \cr & \therefore \,\,x = 6\,g\,;\,{\text{Moles of}}\,C = \frac{6}{{12}} = 0.5 \cr} $$

Specific volume ( volume of $$1$$ $$g$$ ) of cylindrical virus particle $$ = 6.02 \times {10^{ - 2}}\,cc/g$$
Radius of virus $$\left( r \right) = 7\mathop {\text{A}}\limits^{\text{o}} = 7 \times {10^{ - 8}}\,cm$$
Length of virus $$ = 10 \times {10^{ - 8}}\,cm$$
Volume of virus $$ = \pi {r^2}1 = \frac{{22}}{7} \times {\left( {7 \times {{10}^{ - 8}}} \right)^2} \times 10 \times {10^{ - 8}}$$
$$ = 154 \times {10^{ - 23}}\,cc$$
$$Wt.$$  of one virus particle $$ = \frac{{{\text{volume}}}}{{{\text{specific volume}}}}$$
$$\eqalign{ & \therefore \,\,Mol.\,wt.\,{\text{of virus}} = Wt.\,{\text{of}}\,\,{{\text{N}}_A}\,{\text{particle}} \cr & {\text{ = }}\frac{{154 \times {{10}^{ - 23}}}}{{6.02 \times {{10}^{ - 2}}}} \times 6.02 \times {10^{23}} \cr & = 15400\,g/mol \cr & = 15.4\,kg/mol \cr} $$

$$\eqalign{ & {\text{Molality}}\left( m \right) \cr & = \frac{{{\text{No}}{\text{. of moles of solute}}}}{{{\text{Mass}}\,\,{\text{of}}\,\,{\text{solvent}}\,\,\left( {{\text{in}}\,\,kg} \right)}} \cr & {\text{Molality}} \cr & = \frac{{{W_B} \times 1000}}{{{M_B} \times {W_A}\left( {{\text{in}}\,\,g} \right)}} \cr & = \frac{{18.25 \times 1000}}{{36.5 \times 500}} \cr} $$
$$\,\,\, = 1\,m$$     $$\left( {\because \,\,{\text{Molar mass of }}HCl = 36.5\,g\,mo{l^{ - 1}}} \right)$$