9.000$$\,\,m$$  is the most accurate measurement as it is most precise among the given options.

Let atomic weight of other isotope is $$M$$
$$\eqalign{ & 114.82 = \frac{{114.9041 \times 95.72 + M \times 4.28}}{{100}} \cr & M = 112.94 \cr} $$

\[\underset{\begin{smallmatrix} 1\,mole \\ \left( 100\,g \right) \end{smallmatrix}}{\mathop{CaC{{O}_{3}}}}\,+\underset{\begin{smallmatrix} 2\,mole \\ \left( 73\,g \right) \end{smallmatrix}}{\mathop{2HCl}}\,\to CaC{{l}_{2}}+{{H}_{2}}O+C{{O}_{2}}\]
$$100\,g$$  of $$CaC{O_3}$$  reacts with $$73\,g$$  of $$HCl$$
$$40\,g$$  of $$CaC{O_3}$$  will react with $$\frac{{73}}{{100}} \times 40 = 29.2\,g$$     of $$HCl$$
Since $$CaC{O_3}$$  is completely consumed and some amount $$\left( {40 - 29.2 = 10.8\,g} \right)$$    of $$HCl$$  remains unreacted and hence, $$CaC{O_3}$$  is limiting reagent.

\[\underset{\underset{\begin{align} & \\ & \\ & \frac{10}{2}\,mol \\ & \\ & 5\,mol \\ \end{align}}{\mathop{\begin{smallmatrix} \\ \\ 1\,mol \end{smallmatrix}}}\,}{\mathop{{{H}_{2}}}}\,+\underset{\underset{\begin{align} & \\ & \frac{64}{32}mol \\ & \\ & 2\,mol \\ \end{align}}{\mathop{\begin{smallmatrix} \\ \frac{1}{2}mol \end{smallmatrix}}}\,\,}{\mathop{\frac{1}{2}{{O}_{2}}}}\,\to \underset{\underset{\begin{align} & \\ & \\ & \\ & ? \\ \end{align}}{\mathop{\begin{smallmatrix} \,\, \\ \\ 1\,mol \end{smallmatrix}}}\,}{\mathop{{{H}_{2}}O}}\,\]
$$\eqalign{ & \because \,\,\frac{1}{2}mole\,{\text{of}}\,{O_2}\,{\text{gives}} \cr & = 1\,mole\,{\text{of}}\,{H_2}O \cr & \therefore \,\,2\,moles\,{\text{of}}\,{O_2}\,{\text{will}}\,{\text{give}} \cr & = 1\, \times 2 \times 2 \cr & = 4\,moles\,{\text{of}}\,{\text{water}} \cr} $$

$${\text{TIPS/Formulae:}}$$
$$\eqalign{ & {\text{Atomic weight in gms}}\,{\text{ = }}\,{\text{6}}{\text{.023}} \times {\text{1}}{{\text{0}}^{{\text{23}}}}\,{\text{atoms}}\,{\text{ = 1}}\,{\text{Mole}}\,{\text{atoms}} \cr & \left( {\text{i}} \right){\text{Number}}\,{\text{of}}\,{\text{atoms}}\,{\text{in}}\,{\text{24}}g\,{\text{of}}\,C \cr & {\text{ = }}\frac{{24}}{{12}} \times 6.023 \times {10^{23}} = 2 \times 6.023 \times {10^{23}}\,{\text{atoms}} \cr & {\text{ = 2}}\,{\text{mole}}\,{\text{atoms}} \cr & \left( {{\text{ii}}} \right)\,{\text{Number}}\,{\text{of}}\,{\text{atoms}}\,{\text{in}}\,{\text{56}}g\,{\text{of}}\,Fe \cr & {\text{ = }}\frac{{56}}{{56}} \times 6.023 \times {10^{23}} = 6.023 \times {10^{23}}{\text{atom}} \cr & = 1\,{\text{mole}}\,{\text{atoms}} \cr & \left( {{\text{iii}}} \right)\,{\text{Number}}\,{\text{of}}\,{\text{atoms}}\,{\text{in}}\,{\text{27g}}\,{\text{of}}\,Al \cr & {\text{ = }}\frac{{27}}{{27}} \times 6.023 \times {10^{23}} = 6.023 \times {10^{23}}{\text{atom}} \cr & {\text{ = 1}}\,{\text{mole}}\,{\text{atoms}} \cr & \left( {{\text{iv}}} \right)\,{\text{Number}}\,{\text{of}}\,{\text{atoms}}\,{\text{in}}\,{\text{108}}g\,{\text{of}}\,Ag \cr & {\text{ = }}\frac{{108}}{{108}} \times 6.023 \times {10^{23}} = 6.023 \times {10^{23}}\,{\text{atom}} \cr & = 1\,{\text{mole}}\,{\text{atoms}} \cr & \therefore \,24g\,{\text{of}}\,C\,{\text{has}}\,{\text{maximum}}\,{\text{number}}\,{\text{of}}\,{\text{atoms}}. \cr} $$

Zeros between two non-zero digits are significant.

$$\eqalign{ & \left( {\text{A}} \right):Zn + 2HCl \to ZnC{l_2} + {H_2} \cr & 1{\text{ }}mole{\text{ of }}Zn{\text{ produces }}2{\text{ }}g{\text{ of }}{H_2} \cr} $$
$$0.5{\text{ }}mole{\text{ of }}Zn{\text{ will produce}}$$      $$1{\text{ }}g{\text{ of }}{H_2}.$$
$$\eqalign{ & \left( {\text{B}} \right):{\text{Molar mass of}}\,\,{C_{70}}{H_{22}} = 862 \cr & {\text{Mass of atoms}} = \frac{{862}}{{6.023}} \times {10^{23}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1.43 \times {10^{ - 21}}g \cr} $$
$$\left( {\text{C}} \right):71\,g\,\,{\text{of}}\,\,C{l_2}$$    $$ = 6.023 \times {10^{23}}\,{\text{molecules}}$$
$${\text{35}}{\text{.5}}\,{\text{g}}\,\,{\text{of}}\,\,C{l_2} = 3.01 \times {10^{23}}\,{\text{molecules}}$$
$$\eqalign{ & \left( {\text{D}} \right):{\text{Molar mass of}}\,\,S{O_2} = 64 = 1\,mole \cr & 64\,g\,\,{\text{of}}\,\,S{O_2} = 6.023 \times {10^{23}}\,{\text{molecules}} \cr} $$

TIPS/Formulae:
(i) Mass of one electron $$ = 9.108 \times {10^{ - 31}}{\text{kg}}$$
(ii) $$1$$ mole of electron $$ = 6.023 \times {10^{23}}$$   electrons Weight of $$1$$ mole of electron
= Mass of one electron x Avogadro Number
$$ = 9.108 \times {10^{ - 31}} \times 6.023 \times {10^{23}}{\text{kg}}$$
∴ No. of moles of electrons in $$1$$ kg
$$\eqalign{ & {\text{ = }}\frac{1}{{9.108 \times {{10}^{ - 31}} \times 6.023 \times {{10}^{23}}}} \cr & = \frac{1}{{9.108 \times 6.023}} \times {10^8} \cr} $$

On decomposition, $$BaC{O_3}$$  liberates $$C{O_2}$$  as
$$\mathop {BaC{O_3}}\limits_{197\,g} \to BaO\,\,\, + \mathop {C{O_2} \uparrow }\limits_{22.4\,L\,\,\,{\text{at}}\,\,\,STP} $$
$$\because \,197\,g\,$$  of $$BaC{O_3}$$  gives $$ = 22.4\,L$$   of $$C{O_2}$$  at $$STP$$
$$\therefore \,\,9.85\,g$$   of $$BaC{O_3}$$  will give $${\text{ = }}\frac{{22.4 \times 9.85}}{{197}} = 1.12L$$

According to law of conservation of mass, total mass of elements in reactants is equal to the total mass of elements in products.