For the gas $$X$$ ratio of $$\frac{{{C_p}}}{{{C_v}}} = 1:4$$
So, the gas $$X$$ is diatomic.
At $$NTP,$$  volume of $${\text{ }}1{\text{ }}mole{\text{ }}$$  of a gas $$ = 22.4L$$
$$1{\text{ }}mole$$   of a gas $$ = 6.023 \times {10^{23}}{\text{molecules}}$$
Thus, at $$NTP$$  $$22.4{\text{ }}L$$  contains $$ = 6.023 \times {10^{23}}{\text{molecules}}$$
So, at $$NTP{\text{ }}11.2{\text{ }}L$$    contains
$$\eqalign{ & = \frac{{6.023 \times {{10}^{23}} \times 11.2}}{{22.4}}{\text{molecules}} \cr & = 3.01 \times {10^{23}}{\text{molecules}} \cr} $$
Hence, number of atoms of gas $$'X'$$ (diatomic)
$$\eqalign{ & = 3.01 \times {10^{23}} \times 2\,atoms \cr & = 6.02 \times {10^{23}}\,atoms \cr} $$

$$\eqalign{ & \left( {\text{A}} \right)\,1\,{\text{mol}}\,{\text{of}}\,He = 4\,g = {N_A}\,{\text{atoms}} \cr & {\text{or}}\,\,4\,g\,\,{\text{of}}\,\,He = \frac{4}{4}{\text{mol}} = 1\,{N_A}\,{\text{atom}} \cr} $$
$$\left( {\text{B}} \right)\,1\,{\text{mole}}\,\,{\text{of}}\,\,Na = 23\,g = {N_A}\,{\text{atoms}}$$       $$\left[ {\because \,\,{\text{no}}{\text{. of moles}} = \frac{{{\text{give}}\,\,{\text{mass}}}}{{{\text{atomic}}\,\,{\text{mass}}}}} \right]$$
$$\therefore \,46\,g\,\,{\text{of}}\,\,Na = \frac{{46}}{{23}}{\text{mol}}$$     $$ = \frac{{46}}{{23}}\,\,{N_A}\,{\text{atoms}} = 2\,{N_A}\,{\text{atoms}}$$
$$\left( {\text{C}} \right)\,1\,{\text{mole}}\,\,{\text{of}}\,\,Ca = 40\,g = {N_A}\,{\text{atoms}}$$
$$\therefore \,\,0.40\,g\,\,{\text{of}}\,\,Ca = \frac{{0.40}}{{40}}{\text{mol}}$$       $$ = \frac{{0.40}}{{40}}{N_A}\,{\text{atoms}} = 0.01\,{N_A}\,{\text{atoms}}$$
$$\left( {\text{D}} \right)\,1\,{\text{mole}}\,\,{\text{of}}\,\,He = 4\,g = {N_A}\,{\text{atoms}}$$
$$\therefore \,\,12\,g\,\,{\text{of}}\,\,He = \frac{{12}}{4}{\text{mol}}$$     $$ = \frac{{12}}{4}{N_A}\,{\text{atoms}} = 3\,{N_A}\,{\text{atoms}}$$

$$\eqalign{ & {\text{Mass per cent of }}X \cr & = \frac{{{\text{Mass of }}X}}{{{\text{Mass of solution}}}} \times 100 \cr & = \frac{5}{{5 + 45}} \times 100 \cr & = 10\% \cr} $$

$$0.0018\,mL = 0.0018\,g$$     $$ = 0.0001\,mole\,\,{\text{of water}}$$
$$\eqalign{ & {\text{ = 1}}{{\text{0}}^{ - 4}}\,mole \cr & {\text{Number of water molecules}} \cr & = 6.023 \times {10^{23}} \times {10^{ - 4}} \cr & = 6.023 \times {10^{19}} \cr} $$

$$\eqalign{ & 2 + 2\left( {2 + x - 4} \right) = 0\,\left[ {\because Ba{{\left( {{H_2}P{O_2}} \right)}_2}\,{\text{is}}\,{\text{neutral}}\,{\text{molecule}}} \right] \cr & {\text{or}}\,2x - 2 = 0 \Rightarrow x = + 1 \cr} $$

Answer to a multiplication or division is rounded off to the same number of significant figures as possessed by the least precise term in the calculation.

$$\eqalign{ & \frac{{{\text{mass of}}\,A{l_2}{{\left( {S{O_4}} \right)}_3}}}{{{\text{mass of water}}}} \times {10^6} \cr & = 34.2 \cr} $$
1 litre solution contains $$1000\,g$$  of water $$ \Rightarrow $$ In 1 litre solution, mass of
$$\eqalign{ & A{l_2}{\left( {S{O_4}} \right)_3} = \frac{{34.2 \times 1000}}{{{{10}^6}}} = 34.2\,mg \cr & {\text{molarity of}} \cr & A{l_2}{\left( {S{O_4}} \right)_3} = \frac{{34.2 \times {{10}^{ - 3}}}}{{342}}M = {10^{ - 4}}M \cr & A{l_2}{\left( {S{O_4}} \right)_3}\left( {aq} \right) \to 2A{l^{3 + }}\left( {aq} \right) + 3SO_4^{2 - }\left( {aq} \right) \cr & {10^{ - 4}}M\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2 \times {10^{ - 4}}M\,\,\,\,\,\,\,\,\,\,\,3 \times {10^{ - 4}}M \cr & \left[ {SO_4^{2 - }} \right] = 3 \times {10^{ - 4}}M \cr} $$

According to Avogadro's law “equal volumes of all gases contain equal number of molecules under similar conditions of temperature and pressure.” Thus, if $$1\,L$$  of one gas contains $$N$$ molecules, $$2\,L$$  of any gas under similar conditions will contain $$2\,N$$  molecules.

(i) $${H_3}P{O_3}$$  is dibasic acid as it contains two $$ - OH$$  groups.

(ii) Normality = Molarity × basicity of acid.
(iii) Basicity of $${H_3}P{O_3} = 2$$
∴ Normality $$ = 0.3 \times 2 = 0.6$$

$$\mathop {2Mg}\limits_{2 \times 24} + \mathop {{O_2}}\limits_{2 \times 16} \to \mathop {2MgO}\limits_{2\left( {24 + 16} \right)} $$
$$48\,g$$  of $$Mg$$  requires $$32\,g$$  of $${O_2}$$
$$1\,g$$  of $$Mg$$  requires $$\frac{{32}}{{48}} = 0.66\,g$$   of $${O_2}$$
Oxygen available $$ = 0.5\,g$$
Hence, $${O_2}$$  is limiting reagent.
$$32\,g$$  of $${O_2}$$  reacts with $$48\,g$$  of $$Mg$$
$$0.5\,g$$  of $${O_2}$$  will react with $$\frac{{48}}{{32}} \times 0.5 = 0.75\,g$$     of $$Mg$$
Excess of $$Mg = \left( {1.0 - 0.75} \right) = 0.25\,g$$