$$\eqalign{ & {\text{Number of gram atoms of }}H \cr & = \frac{N}{{{N_A}}} = \frac{{100}}{{6.023 \times {{10}^{23}}}} \cr & = 1.66 \times {10^{ - 22}} \cr & {\text{Mass of hydrogen atoms}} \cr & = 1 \times 1.66 \times {10^{ - 22}} \cr & = 1.66 \times {10^{ - 22}}g \cr} $$

Both $$Y$$ and $$X$$ are neither precise nor accurate as the two values in each of them are not close. With respect to $$X$$ & $$Y,$$ the values of $$Z$$ are close & agree with the true value. Hence, both precise & accurate.

$$3$$ molecules of $${O_2} = 1$$  molecules of $$C{S_2}$$
$$6$$ molecules of $${O_2} = 2$$  molecules of $$C{S_2}$$

Element	%	No. of moles	Molar ratio	Whole no. ratio
$$C$$	54.2	$$\frac{{54.2}}{{12}} = 4.5$$	$$\frac{{4.5}}{{2.3}} = 2$$	2
$$H$$	9.2	$$\frac{{9.2}}{1} = 9.2$$	$$\frac{{9.2}}{{2.3}} = 4$$	4
$$O$$	36.6	$$\frac{{36.6}}{{16}} = 2.3$$	$$\frac{{2.3}}{{2.3}} = 1$$	1

Empirical formula $$ = {C_2}{H_4}O$$

$$\eqalign{ & {X_{Sn}} = \frac{{\frac{{12.5}}{{119}}}}{{\frac{{50}}{{209}} + \frac{{25}}{{207}} + \frac{{12.5}}{{119}} + \frac{{12.5}}{{112}}}} \cr & \,\,\,\,\,\,\,\,\,\,\, = 0.176 \cr} $$

$$\eqalign{ & {\text{Normality}} \cr & = \frac{{{N_1}{V_1} - {N_2}{V_2}}}{{{V_1} + {V_2}}} \cr & = \frac{{0.2 \times 100 - 0.1 \times 100}}{{100 + 100}} \cr & = \frac{{10}}{{200}} \cr & = 0.05N\,NaOH \cr} $$

Plan For the calculation of mass of $$AgCl$$  precipitated, we find mass of $$AgN{O_3}$$  and $$NaCl$$  in equal volume with the help of mole concept.
$$16.9\% $$  solution of $$AgN{O_3}$$  means $$16.9\,g\,\,AgN{O_3}$$   is present in $$100 mL$$  solution.
$$\therefore \,\,8.45\,g\,\,AgN{O_3}$$    will be present in $$50\,mL$$  solution.
Similarly,
$$5.8\,g\,\,NaCl$$   is present in $$100 mL$$  solution
$$\therefore \,\,2.9\,g\,\,NaCl\,$$   is present in $$50 mL$$  solution
$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,AgN{O_3} + NaCl\,\,\,\, \to \,\,\,AgCl + NaN{O_3} \cr & {\text{Initial mole}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{8.45}}{{169.8}}\,\,\,\,\,\,\,\,\,\frac{{2.9}}{{58.5}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.049\,\,\,\,\,\, = 0.049 \cr & {\text{After reaction}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.049\,\,\,\,\,\,\,\,\,\,\,0.049 \cr} $$
∴ Mass of $$AgCl$$  precipitated
$$\eqalign{ & = 0.049 \times 143.5 \cr & = 7g \cr} $$

$$\eqalign{ & {\text{Molarity}} \cr & = \frac{{{\text{wt}}{\text{. of solute}}}}{{{\text{Mol}}{\text{.wt}}{\text{. of solute}}}} \times \frac{{100}}{{{\text{Volume of soln}}{\text{.}}\left( {mL} \right)}} \cr & = \frac{{49}}{{98}} \times \frac{{1000}}{{250}} \cr & = 2\,M \cr} $$

$$2\,g\,\,{\text{of}}\,\,{{\text{H}}_2} = 1\,mole,$$     $$32\,g\,\,{\text{of}}\,\,{O_2} = 1\,mole$$
$${\text{Total volume of }}2{\text{ }}moles{\text{ of}}$$      $${\text{gases at }}NTP$$
$$\eqalign{ & = 2 \times 22.4\,L \cr & = 44.8\,L \cr} $$

$${\text{Density}} = \frac{{{\text{Mass}}}}{{{\text{Volume}}}} = \frac{{18.72}}{{1.81}} = 10.34\,g\,c{m^{ - 3}}$$
Since volume contains 3 significant figures hence, answer will be $$10.3\,g\,c{m^{ - 3}}.$$