Laws of chemical combination Questions in English

(A) The law of multiple proportions states that when two elements combine with each other to form more than one compound,the masses of one element that combine with a fixed mass of the other are in a ratio of small whole numbers.
In the case of $CO$ and $CO_2$,carbon $(12.01 \ g)$ combines with oxygen in the ratio of $16.00 \ g$ $(CO)$ and $32.00 \ g$ $(CO_2)$.
The ratio of the masses of oxygen is $16:32$,which simplifies to $1:2$,a simple whole number ratio. Thus,$CO$ and $CO_2$ illustrate the law of multiple proportions.

(D) In the first oxide: $0.5 \ g$ of $A$ combines with $(1.0 - 0.5) = 0.5 \ g$ of oxygen.
So,$1 \ g$ of $A$ combines with $1 \ g$ of oxygen.
In the second oxide: $1.6 \ g$ of $A$ combines with $(4.0 - 1.6) = 2.4 \ g$ of oxygen.
So,$1 \ g$ of $A$ combines with $(2.4 / 1.6) = 1.5 \ g$ of oxygen.
The ratio of the masses of oxygen that combine with a fixed mass $(1 \ g)$ of $A$ is $1 : 1.5$,which is $2 : 3$.
Since this ratio is a simple whole number ratio,the data illustrate the law of multiple proportions.

(C) The law of multiple proportions states that when two elements combine to form more than one compound,the masses of one element that combine with a fixed mass of the other are in the ratio of small whole numbers.
In the pair $CuO$ and $Cu_2O$,copper $(Cu)$ and oxygen $(O)$ combine to form two different compounds.
In $CuO$,$1$ atom of $Cu$ combines with $1$ atom of $O$.
In $Cu_2O$,$2$ atoms of $Cu$ combine with $1$ atom of $O$.
For a fixed mass of oxygen,the ratio of the masses of copper in $CuO$ and $Cu_2O$ is $1:2$,which is a simple whole number ratio.

(A) The percentage of copper and oxygen in a sample of $CuO$ obtained from different methods were found to be the same.
This proves the Law of Constant Proportions because the mass ratio of $Cu : O$ in the compound remains constant regardless of the source or method of preparation.

(C) When two elements combine to form more than one compound,the masses of one element that combine with a fixed mass of the other element are in the ratio of small whole numbers.
In this case,the oxygen content is fixed in the lead oxide samples,and the lead masses are in a simple ratio of $1:2$.
This behavior is described by the Law of Multiple Proportions.

(C) When we balance a chemical equation,we ensure that the number of atoms of each element is the same on both the reactant and product sides.
Since the total number of atoms of each element remains constant,the total mass of the reactants must equal the total mass of the products.
This is in accordance with the $Law$ of $Conservation$ of $Mass$,which states that matter can neither be created nor destroyed in a chemical reaction.

(B) As per the law of multiple proportions,when two elements combine to form two or more compounds,the ratios of the masses of two interacting elements in the two compounds are small whole numbers.
Thus,different proportions of oxygen in the various oxides of nitrogen prove the law of multiple proportion.

(B) The Law of Multiple Proportions states that when two elements combine to form more than one compound,the masses of one element that combine with a fixed mass of the other are in the ratio of small whole numbers.
Given that for a fixed mass of $X$,the masses of $Y$ in compounds $A, B, C, D, E$ are in the ratio $1 : 2 : 3 : 4 : 5$.
In compound $A$,$28$ parts of $X$ combine with $16$ parts of $Y$.
Since the ratio of $Y$ for a fixed amount of $X$ in compound $C$ is $3$ times that of compound $A$,the mass of $Y$ in compound $C$ is $16 \times 3 = 48$ parts.
Therefore,compound $C$ contains $28$ parts of $X$ and $48$ parts of $Y$.

(A) According to the Law of Multiple Proportions,when two elements combine with each other to form two or more than two compounds,the masses of one of the elements that combine with a fixed mass of the other bear a simple whole number ratio.
In $CO$,$12$ parts by mass of carbon combine with $16$ parts by mass of oxygen.
In $CO_2$,$12$ parts by mass of carbon combine with $32$ parts by mass of oxygen.
The ratio of masses of oxygen that combine with a fixed mass of carbon $(12 \ g)$ in these compounds is $16:32$ or $1:2$,which is a simple whole number ratio.
This confirms the Law of Multiple Proportions.

(B) According to the Law of Conservation of Mass,the total mass of the reactants must be equal to the total mass of the products in a chemical reaction.
Given the reaction: $X + Y \rightarrow R + S$.
Mass of reactants = $n + m$.
Mass of products = $p + q$.
Therefore,the relation is $n + m = p + q$.

(A) The correct option is $A$.
$12 \ g$ of carbon combines with $32 \ g$ of oxygen to form $44 \ g$ of carbon dioxide.
$12 \ g + 32 \ g = 44 \ g$.
According to the law of conservation of mass,the total mass of reactants must be equal to the total mass of products.

(D) The law of multiple proportions states that when two elements combine to form more than one compound,the masses of one element that combine with a fixed mass of the other are in the ratio of small whole numbers.
In the case of $SO_2$ and $SO_3$,the element $sulphur$ $(S)$ combines with $oxygen$ $(O)$.
For a fixed mass of $S$ $(32 \ g)$,the mass of $O$ in $SO_2$ is $32 \ g$ and in $SO_3$ is $48 \ g$.
The ratio of the masses of $oxygen$ is $32:48$,which simplifies to $2:3$,a simple whole number ratio.

(B) According to the question:
Compound $A \Rightarrow 1.00 \, g$ of $N$ combines with $0.57 \, g$ of $O$.
Compound $B \Rightarrow 2.00 \, g$ of $N$ combines with $2.24 \, g$ of $O$. Thus,$1.00 \, g$ of $N$ combines with $\frac{2.24}{2} = 1.12 \, g$ of $O$.
Compound $C \Rightarrow 3.00 \, g$ of $N$ combines with $5.11 \, g$ of $O$. Thus,$1.00 \, g$ of $N$ combines with $\frac{5.11}{3} \approx 1.70 \, g$ of $O$.
The ratio of masses of oxygen that combine with a fixed mass $(1.00 \, g)$ of nitrogen in compounds $A, B,$ and $C$ is $0.57 : 1.12 : 1.70$,which simplifies to approximately $1 : 2 : 3$.
Since the ratio is a simple whole number,these results obey the Law of Multiple Proportions.

(A) According to the Law of Reciprocal Proportions,if two elements $C$ and $O$ combine separately with a third element $H$,then the ratio of their masses in which they combine with a fixed mass of $H$ is either the same or a simple multiple of the ratio in which they combine with each other.
In $H_2O$,$2 \ g$ of $H$ combines with $16 \ g$ of $O$.
In $CH_4$,$2 \ g$ of $H$ combines with $6 \ g$ of $C$.
Therefore,the ratio of masses of $C$ and $O$ that combine with a fixed mass of $H$ $(2 \ g)$ is $6 : 16$.
Simplifying $6 : 16$,we get $3 : 8$.

(C) The law of reciprocal proportions states that if two different elements combine separately with a fixed mass of a third element,the ratio of the masses in which they do so are either the same as or a simple multiple of the ratio of the masses in which they combine with each other.
In the given data:
$1$. Hydrogen $(H)$ combines with Oxygen $(O)$ to form $H_2O$: $2 \ g$ of $H$ with $16 \ g$ of $O$.
$2$. Hydrogen $(H)$ combines with Carbon $(C)$ to form $CH_4$: $2 \ g$ of $H$ with $6 \ g$ of $C$.
$3$. Carbon $(C)$ combines with Oxygen $(O)$ to form $CO_2$: $12 \ g$ of $C$ with $32 \ g$ of $O$,which means $6 \ g$ of $C$ combines with $16 \ g$ of $O$.
The ratio of masses of $C$ and $O$ combining with a fixed mass of $H$ $(2 \ g)$ is $6:16$ or $3:8$.
This ratio is the same as the ratio in which $C$ and $O$ combine with each other ($6 \ g$ of $C$ with $16 \ g$ of $O$).
Thus,this illustrates the law of reciprocal proportions.

(D) Let the mass of the element be $100 \ g$ in both cases.
In the first oxide,mass of oxygen $= 53.33 \ g$,so mass of element $= 100 - 53.33 = 46.67 \ g$.
In the second oxide,mass of oxygen $= 36.36 \ g$,so mass of element $= 100 - 36.36 = 63.64 \ g$.
Calculate the mass of oxygen combined with $1 \ g$ of the element:
For oxide $1$: $\frac{53.33}{46.67} \approx 1.143 \ g$.
For oxide $2$: $\frac{36.36}{63.64} \approx 0.571 \ g$.
The ratio of the masses of oxygen that combine with a fixed mass of the element is $1.143 : 0.571 \approx 2 : 1$.
Since the ratio is a simple whole number,this illustrates the Law of Multiple Proportions.

(C) According to the $Law$ of $Conservation$ of $Mass$,matter can neither be created nor destroyed in a chemical reaction.
Therefore,the total mass of the reactants must be equal to the total mass of the products.
Thus,the total mass remains unchanged.

(A) The law of constant composition (also known as the law of definite proportions) states that a given chemical compound always contains its component elements in a fixed ratio by mass,regardless of its source or method of preparation.
Since carbon dioxide $(CO_2)$ always contains $27.27\%$ carbon and $72.73\%$ oxygen by mass regardless of its source,this data supports the Law of constant composition.

(C) The law of definite proportions states that a given chemical compound always contains its component elements in a fixed ratio by mass.
Nitrogen forms multiple oxides with oxygen (e.g.,$N_2O, NO, N_2O_3, NO_2, N_2O_5$).
Since nitrogen exhibits variable valency,it forms different compounds with oxygen,meaning the mass ratio of nitrogen to oxygen is not fixed for the general term 'nitrogen oxide'.
Therefore,the law of definite proportions is not applicable to the generic term 'nitrogen oxide' as it does not refer to a single specific compound.

(B) $H_2O$ contains $H$ and $O$ in a fixed ratio by mass.
This illustrates the law of constant composition,which states that a chemical compound always contains the same elements combined in a fixed proportion by mass.

(B) According to the law of constant proportions,the composition of a compound remains fixed.
Given that $100\, g$ of zinc sulphate crystals contain $22.65\, g$ of zinc.
Therefore,$1\, g$ of zinc sulphate crystals contains $\frac{22.65}{100}\, g$ of zinc.
For $20\, g$ of crystals,the weight of zinc required is calculated as:
$\text{Weight of zinc} = \frac{22.65}{100} \times 20\, g = 4.53\, g$.

(C) In chemical reactions,atoms of different elements combine in fixed ratios to form chemical compounds.
These compounds are held together by chemical bonds,such as ionic or covalent bonds.
Therefore,the correct answer is $C$ (Compounds).

(C) Experiment $I$: Mass of copper oxide $= 2.70 \ g$.
Mass of oxygen $= 2.70 - 2.16 = 0.54 \ g$.
The ratio of masses of copper to oxygen is $2.16 : 0.54 = 4 : 1$.
Experiment $II$: Mass of copper oxide $= 1.15 \ g$.
Mass of copper $= 0.92 \ g$.
Mass of oxygen $= 1.15 - 0.92 = 0.23 \ g$.
The ratio of masses of copper to oxygen is $0.92 : 0.23 = 4 : 1$.
Since the ratio of the masses of copper and oxygen is the same $(4 : 1)$ in both experiments,this illustrates the Law of Definite Proportions.

(C) For compound $A$,the ratio of $Sn$ to $O$ is $78.77 / 21.23 \approx 3.71$.
For compound $B$,the ratio of $Sn$ to $O$ is $88.12 / 11.88 \approx 7.42$.
Now,consider the ratio of the masses of $Sn$ that combine with a fixed mass of $O$ in both compounds:
Ratio $= 3.71 / 7.42 = 1 / 2$.
Since the ratio of the masses of $Sn$ is a simple whole number ratio $(1:2)$,this illustrates the Law of Multiple Proportions.

(B) According to the $Law \ of \ Conservation \ of \ Mass$,the total mass of the reactants must be equal to the total mass of the products in a chemical reaction.
Given the reaction $X + Y \rightarrow R + S$,the total mass of reactants is $(n_1 + n_2) \ g$.
The total mass of products is $(m_1 + m_2) \ g$.
Therefore,the relationship is $n_1 + n_2 = m_1 + m_2$.

(C) In black oxide,$79.9 \ g$ of copper combines with $(100 - 79.9) = 20.1 \ g$ of oxygen.
In red oxide,$88.8 \ g$ of copper combines with $(100 - 88.8) = 11.2 \ g$ of oxygen.
For $79.9 \ g$ of copper in red oxide,the amount of oxygen is $\frac{11.2 \times 79.9}{88.8} \approx 10.08 \ g$.
The ratio of oxygen masses combining with a fixed mass of copper $(79.9 \ g)$ is $20.1 : 10.08 \approx 2 : 1$.
Since the ratio is a simple whole number,this illustrates the Law of Multiple Proportions.

(C) Statement $1$ is based on the Law of Conservation of Mass,which states that atoms are neither created nor destroyed in a chemical reaction. Thus,Statement $1$ is true.
Statement $2$ contradicts Avogadro's Law,which states that equal volumes of all gases under the same conditions of temperature and pressure contain an equal number of molecules. Thus,Statement $2$ is false.

(A) Statement $1$ is true because water $(H_2O)$ always contains hydrogen and oxygen in a fixed mass ratio of $2:16$,which simplifies to $1:8$,regardless of the source.
Statement $2$ is true because Joseph Proust formulated the Law of Definite Proportions in $1799$,which states that a chemical compound always contains exactly the same proportion of elements by mass.
Since Statement $1$ is a direct application of the Law of Definite Proportions (Statement $2$),Statement $2$ is the correct explanation for Statement $1$.

(A) In $H_2O_2$,hydrogen is $5.93\%$,so oxygen is $100 - 5.93 = 94.07 \ g$.
In $H_2O$,hydrogen is $11.2\%$,so oxygen is $100 - 11.2 = 88.8 \ g$.
For a fixed mass of oxygen (say $1 \ g$):
In $H_2O_2$,$1 \ g$ of oxygen combines with $\frac{5.93}{94.07} = 0.063 \ g$ of hydrogen.
In $H_2O$,$1 \ g$ of oxygen combines with $\frac{11.2}{88.8} = 0.126 \ g$ of hydrogen.
The ratio of the masses of hydrogen that combine with a fixed mass of oxygen is $0.063 : 0.126$,which simplifies to $1 : 2$.
Since this is a simple whole-number ratio,the data illustrates the Law of Multiple Proportions.

(D) The Law of Multiple Proportions states that when two elements combine to form more than one compound,the masses of one element that combine with a fixed mass of the other are in the ratio of small whole numbers.
In the pair $SnCl_2$ and $SnCl_4$,the element $Sn$ (Tin) combines with $Cl$ (Chlorine) to form two different compounds.
Here,a fixed mass of $Sn$ combines with different masses of $Cl$ in the ratio $2:4$,which simplifies to $1:2$,a simple whole number ratio.
Therefore,$SnCl_2$ and $SnCl_4$ illustrate the Law of Multiple Proportions.

(B) The reaction between barium chloride and sodium sulfate is given by the equation:
$BaCl_2 + Na_2SO_4 \rightarrow BaSO_4 + 2NaCl$
According to the Law of Conservation of Mass,the total mass of reactants must equal the total mass of products.
Let the mass of $BaCl_2$ be $x \ g$.
Mass of reactants $= (x + 24.4) \ g$
Mass of products $= (46.6 + 23.4) \ g$
Equating both sides:
$x + 24.4 = 46.6 + 23.4$
$x + 24.4 = 70.0$
$x = 70.0 - 24.4$
$x = 45.6 \ g$
Therefore,the weight of $BaCl_2$ required is $45.6 \ g$.

(B) The $Law \ of \ Multiple \ Proportions$ states that when two elements combine to form more than one compound,the masses of one element that combine with a fixed mass of the other are in the ratio of small whole numbers. Nitrogen forms several oxides like $N_2O, NO, N_2O_3, NO_2, N_2O_5$. In these,a fixed mass of nitrogen combines with oxygen in simple whole-number ratios. Therefore,this follows the $Law \ of \ Multiple \ Proportions$.

(C) Gay-Lussac's Law of Gaseous Volumes states that when gases react together,they do so in volumes which bear a simple whole number ratio to one another and to the volumes of the products,if gaseous,provided that the temperature and pressure remain constant. This law is essentially the law of definite proportions expressed in terms of volume.

(A) Moles of $CO_2 = \frac{8.8 \ g}{44 \ g/mol} = 0.2 \ mol$.
$1 \ mol$ of $CO_2$ contains $12 \ g$ of $C$.
So,$0.2 \ mol$ of $CO_2$ contains $0.2 \times 12 = 2.4 \ g$ of $C$.
Moles of $H_2O = \frac{5.4 \ g}{18 \ g/mol} = 0.3 \ mol$.
$1 \ mol$ of $H_2O$ contains $2 \ g$ of $H$.
So,$0.3 \ mol$ of $H_2O$ contains $0.3 \times 2 = 0.6 \ g$ of $H$.
Total mass of $C$ and $H$ in the hydrocarbon $= 2.4 \ g + 0.6 \ g = 3.0 \ g$.
Since the mass of reactants $(3.0 \ g)$ equals the mass of products $(2.4 \ g \ C + 0.6 \ g \ H = 3.0 \ g)$,this follows the Law of Conservation of Mass.

(D) In $NH_3$,$17.65 \ g$ of hydrogen combines with $82.35 \ g$ of nitrogen.
$1 \ g$ of hydrogen combines with $\frac{82.35}{17.65} = 4.67 \ g$ of nitrogen.
In $H_2O$,$11.10 \ g$ of hydrogen combines with $88.90 \ g$ of oxygen.
$1 \ g$ of hydrogen combines with $\frac{88.90}{11.10} = 8.01 \ g$ of oxygen.
The ratio of the masses of $N$ and $O$ that combine with a fixed mass $(1.0 \ g)$ of hydrogen is $4.67 : 8.01 = 1 : 1.72$.
In $NO_2$,the ratio of $N$ and $O$ that combine with each other is $36.85 : 63.15 = 1 : 1.71$.
Since both ratios are the same,the Law of Reciprocal Proportions is illustrated.

(B) The Law of Multiple Proportions was proposed by $John \ Dalton$ in $1803$. This law states that if two elements can combine to form more than one compound,the masses of one element that combine with a fixed mass of the other element are in the ratio of small whole numbers.

(D) According to $Gay-Lussac's$ Law of Gaseous Volumes,when gases react together,they do so in volumes which bear a simple whole number ratio to one another and to the volume of the products,if gaseous,provided that the temperature and pressure remain constant.
In the reaction $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$,the stoichiometric coefficients are $1$,$3$,and $2$ respectively.
Since the reactants and products are in the gaseous state,the ratio of their volumes is $1 : 3 : 2$,which is a simple whole number ratio.
Therefore,this illustrates $Gay-Lussac's$ Law of Gaseous Volumes.

(C) The Law of Multiple Proportions states that when two elements combine to form more than one compound,the masses of one element that combine with a fixed mass of the other are in the ratio of small whole numbers.
In the pair $CuO$ and $Cu_2O$:
$1$. Both compounds are formed by the same two elements: Copper $(Cu)$ and Oxygen $(O)$.
$2$. In $CuO$,$63.5 \ g$ of $Cu$ combines with $16 \ g$ of $O$.
$3$. In $Cu_2O$,$127 \ g$ of $Cu$ $(2 \times 63.5)$ combines with $16 \ g$ of $O$.
$4$. For a fixed mass of $O$ $(16 \ g)$,the ratio of the masses of $Cu$ is $63.5 : 127$,which simplifies to $1 : 2$,a simple whole number ratio.
Therefore,$CuO$ and $Cu_2O$ follow the Law of Multiple Proportions.

(C) Experiment $I$:
$FeO \to Fe + O$
$2.4 \ g \to 1.68 \ g$
Mass of Oxygen $= 2.4 - 1.68 = 0.72 \ g$.
Ratio of $Fe : O = 1.68 : 0.72 = 7 : 3$.
Experiment $II$:
$FeO \to Fe + O$
$2.9 \ g \to 2.03 \ g$
Mass of Oxygen $= 2.9 - 2.03 = 0.87 \ g$.
Ratio of $Fe : O = 2.03 : 0.87 = 7 : 3$.
Since the ratio of the masses of elements in the compound is constant,it follows the Law of Definite Proportions.

(D) The molar mass of $CO_2 = 44 \ g/mol$.
$44 \ g$ of $CO_2$ contains $12 \ g$ of carbon.
$1.32 \ g$ of $CO_2$ contains $(1.32 \times 12) / 44 = 0.36 \ g$ of carbon.
The molar mass of $H_2O = 18 \ g/mol$.
$18 \ g$ of $H_2O$ contains $2 \ g$ of hydrogen.
$2.7 \ g$ of $H_2O$ contains $(2.7 \times 2) / 18 = 0.30 \ g$ of hydrogen.
Total mass of carbon and hydrogen after combustion $= 0.36 \ g + 0.30 \ g = 0.66 \ g$.
Since the total mass of the elements before and after the reaction remains the same,it follows the Law of Conservation of Mass.

(B) In compound $A$,the ratio of $x:y$ is $1:1$. Given $28 \ g$ of $x$ combines with $16 \ g$ of $y$.
In compound $C$,the ratio of $x:y$ is $1:3$.
For a fixed amount of $x$ $(28 \ g)$,the amount of $y$ required for compound $C$ is $16 \times 3 = 48 \ g$.
Thus,$48 \ g$ of $y$ combines with $28 \ g$ of $x$.
Therefore,$24 \ g$ of $y$ will combine with $(28 / 48) \times 24 = 14 \ g$ of $x$.

(B) In the first compound,$C = 42.9\%$ and $O = 57.1\%$.
Ratio of $C:O = \frac{42.9}{57.1} = 0.751:1$.
In the second compound,$C = 27.3\%$ and $O = 72.7\%$.
Ratio of $C:O = \frac{27.3}{72.7} = 0.376:1$.
Comparing the ratio of $C$ for a fixed amount of $O$:
Ratio $= 0.751 : 0.376 = 2 : 1$.
Since the ratio is a simple whole number,this illustrates the Law of Multiple Proportions.

(B) The Law of Multiple Proportions states that when two elements combine to form more than one compound,the masses of one element that combine with a fixed mass of the other are in the ratio of small whole numbers.
In the set $N_2O, N_2O_3, N_2O_5$,nitrogen and oxygen combine to form different compounds.
For a fixed mass of nitrogen $(28 \ g)$,the masses of oxygen are $16 \ g, 48 \ g, \text{ and } 80 \ g$ respectively.
The ratio of these masses is $16:48:80$,which simplifies to $1:3:5$,a ratio of small whole numbers.
Therefore,this set illustrates the Law of Multiple Proportions.

(D) Total mass before the chemical reaction = Mass of $AgNO_3$ + Mass of $NaCl$ + Mass of water
$= 1.7 \, g + 0.585 \, g + 200 \, g = 202.285 \, g$
Total mass after the chemical reaction = Mass of $AgCl$ + Mass of $NaNO_3$ + Mass of water
$= 1.435 \, g + 0.85 \, g + 200 \, g = 202.285 \, g$
Since the total mass before the reaction is equal to the total mass after the reaction,it follows the $Law \, of \, Conservation \, of \, Mass$.

(C) According to the Law of Reciprocal Proportions,if two different elements combine separately with a third element,the ratio in which they do so will be the same or a simple multiple of the ratio in which they combine with each other.
Here,$A$ combines with $B$ in ratio $1:2$ (or $3:6$ by mass).
$A$ combines with $C$ in ratio $6:4$ (or $6:4$ by mass).
For a fixed mass of $A$ ($6$ parts),$B$ and $C$ combine in the ratio $3:4$.
Since $B$ and $C$ combine in a simple whole number ratio,this follows the Law of Reciprocal Proportions.

(C) According to the Law of Definite Proportions,the percentage composition of a compound remains constant regardless of the source.
Given that $Ca$ constitutes $40\%$ of $CaCO_3$ by mass.
For a $4 \ g$ sample of $CaCO_3$,the mass of $Ca$ is calculated as:
$\text{Mass of } Ca = \frac{40}{100} \times 4 \ g = 1.6 \ g$.

(B) $100 \, g$ of $ZnSO_4$ crystals are obtained from $22.65 \, g$ of $Zn$.
$1 \, g$ of $ZnSO_4$ crystals can be obtained from $\frac{22.65}{100} \, g$ of $Zn$.
Therefore,the amount of $Zn$ required to obtain $20 \, g$ of $ZnSO_4$ crystals is $\frac{22.65}{100} \times 20 = 4.53 \, g$.

(B) Experiment $I$:
Mass of $CuO = 5.06 \ g$,Mass of $Cu = 4.04 \ g$.
Mass of $O = 5.06 - 4.04 = 1.02 \ g$.
Ratio of $Cu : O = 4.04 / 1.02 \approx 3.96 : 1$.
Experiment $II$:
Mass of $Cu = 1.3 \ g$,Mass of $CuO = 1.63 \ g$.
Mass of $O = 1.63 - 1.3 = 0.33 \ g$.
Ratio of $Cu : O = 1.3 / 0.33 \approx 3.94 : 1$.
Since the ratio of the masses of elements in the compound is the same in both experiments,it illustrates the Law of Definite Proportions.

(B) The Law of Definite Proportions states that a given chemical compound always contains its component elements in a fixed ratio by mass. In $H_2O$,the mass ratio of $H$ to $O$ is always $2:16$ or $1:8$,regardless of the source of the water.

(A) The molar mass of $SO_2$ is $32 + 32 = 64 \ g/mol$.
At $N.T.P$,$22400 \ mL$ of $SO_2$ weighs $64 \ g$.
Therefore,$224 \ mL$ of $SO_2$ weighs $\frac{224 \times 64}{22400} = 0.64 \ g$.
In the first experiment,$0.64 \ g$ of $SO_2$ contains $0.32 \ g$ of sulfur.
Thus,the percentage of sulfur is $\frac{0.32}{0.64} \times 100 = 50\%$.
In the second experiment,the compound also contains $50\%$ sulfur.
Since the composition of sulfur and oxygen in $SO_2$ remains constant regardless of the source or method of preparation,it follows the Law of Definite Proportions.