In $$XO,50\,g$$   of element combines with $$50\,g$$  of oxygen.
∴ $$1\,g$$  of element combines with $$1\,g$$  of oxygen.
In $$X{O_2},40\,g$$    of element combines with $$60\,g$$  of oxygen.
∴ $$1\,g$$  of element combines with $$1.5\,g$$  of oxygen.
Thus, ratio of masses of oxygen which combines with $$1\,g$$  of element is 1 : 1.5 or 2 : 3. This is in accordance with the law of multiple proportions. In (B) Gay Lussac's law of gaseous volume is followed. In (C) law of conservation of mass is followed while in (D) Avogadro's law is followed.

$$8008 = 8.008 \times {10^3}$$

The relation between molarity $$(M)$$  and molality $$(m)$$  is
$$d = M\left( {\frac{1}{m} + \frac{{{M_2}}}{{1000}}} \right),$$     $${M_2} = Mol.\,\,mass\,\,{\text{of solute}}$$
$$\eqalign{ & {\text{On putting value}} \cr & 1.252 = 3\left( {\frac{1}{m} + \frac{{58.5}}{{1000}}} \right) \cr & {\text{on solving}}\,\,m = 2.79 \cr} $$

$$\eqalign{ & {\text{Weight of gas}} = 0.24\,g \cr & {\text{Volume of gas}}\left( V \right) = 45\,mL = 0.045\,L \cr & {\text{Density of}}\,{H_2}\left( d \right) = 0.089 \cr & {\text{Weight of }}45{\text{ }}mL{\text{ of }}{H_2} = V \times d \cr & = 0.045 \times 0.089 \cr & = 4.005 \times {10^{ - 3}}g \cr & {\text{Therefore, vapour density}} \cr & = \frac{{{\text{Weight of certain volume of substance}}}}{{{\text{Weight of same volume of hydrogen}}}} \cr & = \frac{{0.24}}{{4.005 \times {{10}^{ - 3}}}} \cr & = 59.93 \cr} $$

Element	%	No. of moles	Molar ratio	Whole no. ratio
$$P$$	27.3	$$\frac{{27.3}}{{12}} = 2.27$$	1	1
$$Q$$	72.7	$$\frac{{72.7}}{{16}} = 4.54$$	2	2

Empirical formula $$ = P{Q_2}$$

Physical and chemical properties of a compound are different from those of its constituent elements.

The required equation is
$$\eqalign{ & 2KMn{O_4} + 3{H_2}S{O_4} \to \cr & {K_2}S{O_4} + 2MnS{O_4} + 3{H_2}O + \mathop {5\left[ O \right]}\limits_{{\text{nascent oxygen}}} \cr & 2Fe\left( {{C_2}{O_4}} \right) + 3{H_2}S{O_4} + 3\left[ O \right] \to \cr & \,\,\,\,\,\,\,\,\,\,\,\,F{e_2}{\left( {S{O_4}} \right)_3} + 2C{O_2} + 3{H_2}O \cr} $$
$$O$$  required for $$1\,mol.$$  of $$Fe\left( {{C_2}{O_4}} \right)$$   is 1.5, $$5$$ $$O$$  are obtained from $$2\,moles$$   of $$KMn{O_4}$$
$$\therefore \,\,1.5\,\left[ O \right]$$   will be obtained from
$$ = \frac{2}{5} \times 1.5 = 0.6\,moles$$     of $$KMn{O_4}.$$

$${\text{No}}{\text{. of moles in 0}}{\text{.365}}\,g\,\,{\text{of}}\,\,HCl$$       $$ = \frac{{0.365}}{{36.5}} = 0.01$$
$$\eqalign{ & {\text{Volume of solution in}}\,\,L = \frac{{0.01}}{{1000}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.1\,L \cr & {\text{Molarity}} = \frac{n}{{V\left( {{\text{in}}\,\,L} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{0.01}}{{0.1}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.1\,M \cr} $$

$${\text{1}}\,Mole$$  of $$M{g_3}{\left( {P{O_4}} \right)_2}$$   contains $${\text{8}}\,\,mole$$  of oxygen atoms
$$\therefore \,\,8\,mole$$   of oxygen atoms $$ \equiv \,1\,mole$$   of $$M{g_3}{\left( {P{O_4}} \right)_{2\,}}mole$$    of $$M{g_3}{\left( {P{O_4}} \right)_2}$$
$$0.25\,mole$$   of oxygen atom $$ \equiv \frac{1}{8} \times 0.25\,mole$$    of $$M{g_3}{\left( {P{O_4}} \right)_2}$$
$$ = 3.125 \times {10^{ - 2}}\,mole$$     of $$M{g_3}{\left( {P{O_4}} \right)_2}$$

In first experiment,
Mass of iron oxide $$ = 2.4\,g$$
Mass of iron $$ = 1.68\,g$$
Mass of oxygen $$ = 2.4 - 1.68 = 0.72$$
Ratio of masses 0f Iiron and oxygen $$ = \frac{{1.68}}{{0.72}} = 7:3$$
In second experiment,
Mass of iron oxide $$ = 2.7\,g$$
Mass of iron $$ = 1.89\,g$$
Mass of oxygen $$ = 2.7 - 1.89 = 0.81$$
Ratio of masses of iron and oxygen $$ = \frac{{1.89}}{{0.81}} = 7:3$$
The same ratio confirms that these experiments clarify Law of constant proportions.