1. A sample is prepared by dissolving sugar completely in water. The resulting liquid has the same composition throughout. How should this sample be classified?
ⓐ. Pure compound
ⓑ. Homogeneous mixture
ⓒ. Heterogeneous mixture
ⓓ. Pure element
Correct Answer: Homogeneous mixture
Explanation: A homogeneous mixture contains two or more substances distributed uniformly throughout the sample. In the given liquid, sugar and water are two chemically distinct substances. Dissolving sugar does not convert these components into a new chemical compound. The sugar particles become uniformly distributed in the water, producing a homogeneous mixture. Uniform composition throughout a particular sample does not necessarily indicate a pure substance. A pure compound has a characteristic fixed chemical composition, whereas the sugar-to-water proportion can be changed. Adding more sugar or water changes the composition without requiring a chemical reaction. Therefore, the correct classification is a homogeneous mixture.
2. A pure liquid compound is converted completely into vapour without undergoing any chemical reaction. Which statement correctly describes this change?
ⓐ. It changes from a compound into a homogeneous mixture.
ⓑ. It changes from a pure substance into a mixture.
ⓒ. Its physical state changes but its chemical classification remains unchanged.
ⓓ. Its constituent elements change although the sample remains pure.
Correct Answer: Its physical state changes but its chemical classification remains unchanged.
Explanation: Matter can be classified independently according to physical state and chemical composition. The initial sample is a liquid, while the final sample is a gas. A liquid has a definite volume but takes the shape of its container. A gas possesses neither a definite shape nor a definite volume and occupies the available space. During an ordinary liquid-to-vapour transition, the chemical identity of the constituent particles remains unchanged. No additional chemical substance is produced merely because the particles become more widely separated. Since the starting material is a pure compound and no chemical reaction occurs, its vapour remains the same pure compound. This distinction is important because physical-state changes and chemical changes are not equivalent processes. Thus, the physical classification changes, whereas the chemical-composition classification does not.
3. Two transparent solutions are prepared separately. Solution P contains 5 g of sodium chloride dissolved in 100 g of water, while solution Q contains 15 g of sodium chloride dissolved in 100 g of water. Both solutions are completely uniform, with no undissolved solid. Which classification is correct?
ⓐ. P is a pure compound, while Q is a homogeneous mixture.
ⓑ. Both P and Q are homogeneous mixtures.
ⓒ. P is a homogeneous mixture, while Q is a heterogeneous mixture.
ⓓ. Both P and Q are pure compounds of different compositions.
Correct Answer: Both P and Q are homogeneous mixtures.
Explanation: A homogeneous mixture has its components uniformly distributed throughout the sample. Both solutions contain sodium chloride and water as distinct chemical substances. Their components are present in different proportions, but this does not change the basic classification. Sodium chloride dissolves in water without becoming a new compound having a fixed sodium-chloride-to-water ratio. Solution P contains less dissolved salt relative to water than solution Q. Nevertheless, both are uniform because the question specifies complete dissolution and absence of undissolved solid. A pure compound cannot acquire a different characteristic elemental composition merely by changing the quantities used to prepare a sample. By contrast, variable composition is a defining possibility for mixtures. Therefore, both preparations remain homogeneous mixtures despite their different concentrations.
4. Consider the following statements about the classification of matter.
Statement I: Every sample containing atoms of two different elements must be a mixture.
Statement II: A mixture may have a uniform composition throughout a particular sample.
Statement III: A pure compound may exist in different physical states without changing its chemical identity.
Statement IV: Mixing two pure substances without a chemical reaction necessarily produces a new compound.
Which combination contains only the correct statements?
ⓐ. I and II only
ⓑ. I and IV only
ⓒ. II and IV only
ⓓ. II and III only
Correct Answer: II and III only
Explanation: Statement I is incorrect because a pure compound can contain atoms of two or more different elements chemically combined in a definite proportion. Water, for example, contains hydrogen and oxygen but is a compound rather than a mixture. The presence of different elements must not be confused with the presence of different chemical substances. Statement II is correct because homogeneous mixtures have uniformly distributed components throughout the bulk of the sample. Uniformity in one sample does not imply that its overall composition is fixed for every possible preparation. Statement III is correct because an ordinary change of physical state does not alter the chemical identity of a compound. Liquid water and water vapour contain the same molecular species under conditions where no chemical reaction occurs. Statement IV is incorrect because physical mixing does not necessarily involve the formation of new chemical bonds or a new compound. The original substances can retain their chemical identities in the mixture. Consequently, only Statements II and III are correct.
5. Three containers P, Q and R contain the particles described below. X and Y represent atoms of different elements.
Container P contains only identical particles. Each particle consists of one X atom chemically bonded to one Y atom.
Container Q contains separate X atoms and separate Y atoms. These particles are uniformly distributed throughout the container, but X and Y are not chemically bonded.
Container R contains only identical particles. Each particle consists of two chemically bonded X atoms.
Which classification of P, Q and R is correct?
ⓐ. P—pure compound; Q—homogeneous mixture; R—pure element
ⓑ. P—homogeneous mixture; Q—pure compound; R—pure element
ⓒ. P—pure compound; Q—pure element; R—homogeneous mixture
ⓓ. P—pure element; Q—homogeneous mixture; R—pure compound
Correct Answer: P—pure compound; Q—homogeneous mixture; R—pure element
Explanation: The classification must be based on the chemical identities and combinations of the constituent particles. In container P, every particle contains one X atom chemically combined with one Y atom. All particles have the same composition, and the two different elements occur in a fixed atomic ratio. Therefore, P represents a pure compound. Container Q contains two chemically different kinds of particles that are not bonded to one another. Their uniform distribution makes Q a homogeneous mixture rather than a compound. The mere presence of both elements in a container does not establish chemical combination. Container R contains particles made exclusively from X atoms. Although each particle contains two atoms, both atoms belong to the same element. Consequently, R represents a pure elemental substance. The decisive distinction is between different atom types within one chemically defined particle and different chemical substances coexisting in a sample.
6. A sealed container contains only \(\mathrm{H_2O}\), present simultaneously as liquid water and water vapour. No air or other chemical substance is present. Which description is correct?
ⓐ. A heterogeneous mixture of two compounds
ⓑ. One pure compound present in two physical states
ⓒ. One homogeneous mixture containing two phases
ⓓ. Two distinct pure compounds in separate phases
Correct Answer: One pure compound present in two physical states
Explanation: The chemical identity of the material is determined by its constituent chemical species rather than by the number of physical phases present. Both the liquid and vapour contain \(\mathrm{H_2O}\) molecules. During vaporisation, molecules become more widely separated and their freedom of movement increases. This change does not convert water into a chemically different substance. The liquid and vapour are distinguishable physical phases, so the system is not uniform in physical state throughout the container. However, the presence of two phases does not establish the presence of two different chemical substances. Since only \(\mathrm{H_2O}\) is present, the material remains chemically pure. It is therefore incorrect to classify the container contents as a mixture of two compounds. A sample can exhibit physical phase separation while retaining one definite chemical composition.
7. A pure compound P contains elements X and Y in the masses shown below. Two additional specimens, Q and R, contain only X and Y, but their chemical identities are unknown.
| Specimen | Mass of X | Mass of Y |
|---|
| P (known pure compound) | 4 g | 6 g |
| Q | 8 g | 12 g |
| R | 8 g | 14 g |
Which inference is justified by the composition data?
ⓐ. Q must be pure P; R must be a different pure compound.
ⓑ. Q could be pure P; R could contain P with 2 g of additional Y.
ⓒ. Q must be a mixture; R could contain P with 2 g of additional X.
ⓓ. Q could be pure P; R could contain P with 2 g of additional X.
Correct Answer: Q could be pure P; R could contain P with 2 g of additional Y.
Explanation: \(\textbf{Composition of the known compound:}\)
A pure compound has a characteristic fixed elemental mass ratio.
For specimen P, the ratio is
\[
m_X:m_Y=4:6=2:3
\]
\(\textbf{Examination of specimen Q:}\)
The elemental mass ratio in Q is
\[
m_X:m_Y=8:12=2:3
\]
Thus, Q has a composition compatible with pure compound P.
However, matching elemental mass ratios alone do not prove that two unidentified specimens contain the same chemical compound.
\(\textbf{Examination of specimen R:}\)
If the entire 8 g of X in R belongs to compound P, the required mass of combined Y is
\[
m_Y=8\times\frac{6}{4}
\]
\[
m_Y=12\,\text{g}
\]
The observed mass of Y in R is 14 g.
Hence, the additional amount of Y is
\[
m_{\text{additional Y}}=14-12
\]
\[
m_{\text{additional Y}}=2\,\text{g}
\]
Specimen R could therefore consist of 20 g of P mixed with 2 g of uncombined Y.
This is a possible composition, not proof of the specimen's actual chemical constitution.
\(\textbf{Final inference:}\)
Q is compositionally compatible with P, while R could represent P mixed with additional Y.
The key distinction is between compatibility with fixed composition and proof of chemical identity.
8. Two samples L and M are examined. L is a uniform aqueous glucose solution from which glucose and water can be recovered separately by suitable physical methods. M consists entirely of \(\mathrm{H_2O}\) molecules and can be decomposed into hydrogen and oxygen by a chemical process. Which classification follows?
ⓐ. L is a pure compound; M is a homogeneous mixture.
ⓑ. L is a heterogeneous mixture; M is a pure compound.
ⓒ. L is a homogeneous mixture; M is a pure compound.
ⓓ. L is a homogeneous mixture; M is a mixture of elements.
Correct Answer: L is a homogeneous mixture; M is a pure compound.
Explanation: The two samples must be distinguished according to the nature of their constituent substances and the changes needed for separation. Sample L contains glucose and water as distinct components. The uniform distribution of dissolved glucose makes the sample homogeneous. Glucose can be recovered using an appropriate physical separation process without converting its constituent elements into new substances. Therefore, L is a homogeneous mixture. Sample M contains identical \(\mathrm{H_2O}\) molecules and has a definite chemical composition. Hydrogen and oxygen are chemically combined within those molecules. Separating them into their elemental forms requires a chemical change rather than a simple physical separation. The existence of two different elements within each water molecule does not make M a mixture. Thus, the observations support the classification of L as a homogeneous mixture and M as a pure compound.
9. A hot, homogeneous solution contains a dissolved solid whose solubility decreases on cooling. The solution is cooled until some pure solute crystals separate, while some solute remains dissolved. The crystals are then removed completely by filtration, leaving a uniform liquid filtrate. What is the classification of the material at the three successive stages?
ⓐ. Homogeneous mixture → heterogeneous mixture → homogeneous mixture
ⓑ. Homogeneous mixture → homogeneous mixture → pure compound
ⓒ. Pure compound → heterogeneous mixture → homogeneous mixture
ⓓ. Homogeneous mixture → heterogeneous mixture → pure compound
Correct Answer: Homogeneous mixture → heterogeneous mixture → homogeneous mixture
Explanation: Initially, the dissolved solute is uniformly distributed throughout the solvent, so the system is a homogeneous mixture. Cooling reduces the amount of solute that can remain dissolved under the stated conditions. Some solute consequently separates as solid crystals. The system now contains a solid crystalline phase and a liquid solution phase. Its composition is no longer uniform throughout the complete sample, so the system is heterogeneous at this stage. Crystallisation does not necessarily involve a chemical reaction; the solute retains its chemical identity. Filtration physically separates the crystals from the liquid solution. The question specifies that the filtrate is uniform and still contains dissolved solute. Therefore, the filtrate remains a homogeneous mixture rather than becoming a pure compound. The sequence illustrates how changes in physical conditions and separation can alter mixture homogeneity without requiring formation of a new chemical substance.
10. A homogeneous solution contains 16 g of sodium chloride in a total solution mass of 80 g. A 20 g portion of this solution is removed, after which 40 g of pure water is added to the remaining solution and mixed thoroughly. What is the final mass percentage of sodium chloride?
ⓐ. \(8\%\)
ⓑ. \(16\%\)
ⓒ. \(12\%\)
ⓓ. \(20\%\)
Correct Answer: \(12\%\)
Explanation: \(\textbf{Initial composition:}\)
The solution contains 16 g of sodium chloride in 80 g of solution.
The initial salt mass fraction is
\[
w_{\text{salt}}=\frac{16}{80}=0.20
\]
Thus, the starting solution contains \(20\%\) sodium chloride by mass.
\(\textbf{Salt removed with the solution:}\)
Because the solution is homogeneous, the withdrawn 20 g portion has the same salt mass fraction as the original solution.
The mass of salt withdrawn is
\[
m_{\text{salt, withdrawn}}=20\times0.20
\]
\[
m_{\text{salt, withdrawn}}=4\,\text{g}
\]
\(\textbf{Composition of the residue:}\)
The salt remaining is
\[
m_{\text{salt, remaining}}=16-4
\]
\[
m_{\text{salt, remaining}}=12\,\text{g}
\]
The solution mass remaining is
\[
m_{\text{solution, remaining}}=80-20=60\,\text{g}
\]
\(\textbf{Effect of adding pure water:}\)
Water increases the total mass but does not introduce additional sodium chloride.
The final solution mass becomes
\[
m_{\text{final}}=60+40=100\,\text{g}
\]
The final salt mass percentage is
\[
w_{\text{final}}=\frac{12}{100}\times100
\]
\[
w_{\text{final}}=12\%
\]
\(\textbf{Consistency check:}\)
The salt concentration decreases from \(20\%\) to \(12\%\), as expected after adding pure water.
The final material remains a homogeneous mixture, assuming the salt is completely dissolved.
\(\textbf{Final answer:}\)
\[
\boxed{12\%}
\]
11. Consider the following Assertion and Reason.
Assertion (A): In every ordinary chemical reaction, the number of molecules must remain unchanged because the total mass is conserved.
Reason (R): According to Dalton's atomic theory, chemical reactions involve rearrangement of atoms without their creation or destruction.
Select the correct option.
ⓐ. Both A and R are true, and R correctly explains A.
ⓑ. Both A and R are true, but R does not correctly explain A.
ⓒ. A is true, but R is false.
ⓓ. A is false, but R is true.
Correct Answer: A is false, but R is true.
Explanation: The Assertion is false because conservation of mass does not require conservation of the total number of molecules. A chemical reaction may convert several reactant molecules into a different number of product molecules. For example, in the reaction \(2\mathrm{H_2}+\mathrm{O_2}\rightarrow2\mathrm{H_2O}\), three reactant molecules produce two product molecules when the coefficients are interpreted as particle counts. Nevertheless, four hydrogen atoms and two oxygen atoms occur on each side of the equation. The Reason correctly expresses Dalton's postulate that ordinary chemical reactions involve rearrangement rather than creation or destruction of atoms. Conservation applies separately to the atoms of each element in the reaction. The total mass is therefore conserved in a closed reacting system. Molecular identity and molecular count may change even though the constituent atoms are conserved. This distinction prevents the common error of equating conservation of matter with conservation of the number of molecules. Hence, the Assertion is false and the Reason is true.
12. In a hypothetical chemical reaction, six molecules of \(\mathrm{X_2}\) react completely with four molecules of \(\mathrm{Y_2}\). Exactly four identical molecules of a single compound Z are formed, with no other products and no unreacted starting material. What is the molecular formula of Z?
ⓐ. \(\mathrm{X_2Y_3}\)
ⓑ. \(\mathrm{X_3Y_2}\)
ⓒ. \(\mathrm{X_6Y_4}\)
ⓓ. \(\mathrm{X_3Y_4}\)
Correct Answer: \(\mathrm{X_3Y_2}\)
Explanation: \(\textbf{Atom conservation principle:}\)
Dalton's atomic theory states that atoms are neither created nor destroyed in an ordinary chemical reaction.
Therefore, the total number of X atoms and Y atoms must be conserved separately.
\(\textbf{Counting the initial X atoms:}\)
Each \(\mathrm{X_2}\) molecule contains two X atoms.
Hence,
\[
N_X=6\times2
\]
\[
N_X=12\;\text{atoms}
\]
\(\textbf{Counting the initial Y atoms:}\)
Each \(\mathrm{Y_2}\) molecule contains two Y atoms.
Hence,
\[
N_Y=4\times2
\]
\[
N_Y=8\;\text{atoms}
\]
\(\textbf{Distributing atoms among product molecules:}\)
The reaction produces exactly four identical molecules of Z.
Every X atom must therefore be distributed among these four molecules.
The number of X atoms per molecule is
\[
n_X=\frac{12}{4}=3
\]
Similarly, the number of Y atoms per molecule is
\[
n_Y=\frac{8}{4}=2
\]
Thus, each product molecule contains three X atoms and two Y atoms.
\(\textbf{Verification:}\)
The resulting symbolic equation is
\[
6\mathrm{X_2}+4\mathrm{Y_2}\rightarrow4\mathrm{X_3Y_2}
\]
The reactant side contains 12 X atoms and 8 Y atoms.
The product side also contains 12 X atoms and 8 Y atoms.
Thus, the proposed formula satisfies the required atom conservation.
\(\textbf{Final answer:}\)
\[
\boxed{\mathrm{X_3Y_2}}
\]
13. Two pure compounds are formed from the same hypothetical elements X and Y. Their experimental mass compositions are given below.
| Compound | Mass of X | Mass of Y |
|---|
| P | 3 g | 9 g |
| Q | 5 g | 30 g |
Which inference is consistent with Dalton's explanation of chemical combination?
ⓐ. Both compounds have identical mass ratios and therefore identical atom ratios.
ⓑ. Their relative Y masses are in the ratio 2:1, so both must be the same compound.
ⓒ. The relative Y masses are in the ratio 1:2, consistent with different fixed atom ratios.
ⓓ. The data represent continuously variable composition within one pure compound.
Correct Answer: The relative Y masses are in the ratio 1:2, consistent with different fixed atom ratios.
Explanation: \(\textbf{Comparison condition:}\)
When comparing compounds containing the same elements, the mass of one element must be held fixed.
The original samples contain different masses of X, so their Y masses cannot be compared directly.
\(\textbf{Compound P:}\)
The mass of Y combined per gram of X is
\[
\frac{m_Y}{m_X}=\frac{9}{3}=3
\]
Thus, each gram of X is combined with 3 g of Y in P.
\(\textbf{Compound Q:}\)
For compound Q,
\[
\frac{m_Y}{m_X}=\frac{30}{5}=6
\]
Thus, each gram of X is combined with 6 g of Y in Q.
\(\textbf{Comparison at fixed X mass:}\)
The corresponding masses of Y are therefore in the ratio
\[
m_Y(P):m_Y(Q)=3:6
\]
\[
m_Y(P):m_Y(Q)=1:2
\]
\(\textbf{Chemical interpretation:}\)
Dalton proposed that compounds form when atoms combine in fixed ratios involving whole numbers.
The same two elements may form different compounds when they combine in different fixed atomic proportions.
The observed simple mass relationship is consistent with this interpretation and the law of multiple proportions.
The result does not require either compound to have a continuously variable composition.
It is also incorrect to conclude that these are necessarily the same compound merely because the same elements are present.
A numerical mass ratio alone does not uniquely establish the molecular formula of either compound.
\(\textbf{Final inference:}\)
The masses of Y combining with a fixed mass of X are in the ratio \(1:2\), consistent with distinct fixed combinations of atoms.
14. Consider the following statements about Dalton's atomic theory and modern atomic understanding.
Statement I: Atoms belonging to the same chemical element may possess different atomic masses.
Statement II: Atoms of the same element having different masses must therefore be classified as atoms of different elements.
Choose the correct option.
ⓐ. Statement I is true, but Statement II is false.
ⓑ. Statement I is false, but Statement II is true.
ⓒ. Both Statement I and Statement II are true.
ⓓ. Both Statement I and Statement II are false.
Correct Answer: Statement I is true, but Statement II is false.
Explanation: Statement I is correct because isotopes of a chemical element can have different atomic masses. Dalton's original theory proposed that all atoms of a given element possess identical properties, including identical mass. The discovery of isotopes showed that the identical-mass part of this postulate requires modification. Different isotopes of an element are still atoms of that same chemical element. Consequently, Statement II is incorrect. Difference in atomic mass alone does not establish that two atoms belong to different elements. Modern atomic understanding distinguishes elemental identity from isotopic mass variation. This qualification does not invalidate the usefulness of Dalton's central idea that chemical reactions involve rearrangement of atoms. The historical postulate and the modern refinement must therefore be interpreted separately. Hence, Statement I is true, while Statement II is false.
15. At the same temperature and pressure, 2 L of hydrogen gas react completely with 1 L of oxygen gas to produce 2 L of water vapour. Assuming the gases obey Avogadro's law, what is the ratio of the number of product gas molecules to the total number of reactant gas molecules initially present?
ⓐ. \(\frac{1}{2}\)
ⓑ. \(1\)
ⓒ. \(\frac{3}{2}\)
ⓓ. \(\frac{2}{3}\)
Correct Answer: \(\frac{2}{3}\)
Explanation: \(\textbf{Applicable gas-volume principle:}\)
Avogadro's law states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.
The specified temperature-pressure condition is essential when comparing gas volumes through molecular counts.
\(\textbf{Balanced chemical relationship:}\)
Hydrogen and oxygen exist as diatomic molecules under the stated conditions.
Their reaction to form water vapour is
\[
2\mathrm{H_2(g)}+\mathrm{O_2(g)}\rightarrow2\mathrm{H_2O(g)}
\]
The reaction is balanced because hydrogen and oxygen atom counts are conserved.
\(\textbf{Initial molecular count:}\)
Let \(N\) represent the number of molecules present in 1 L of any of these gases at the given temperature and pressure.
The 2 L hydrogen sample contains
\[
N_{\mathrm{H_2}}=2N
\]
The 1 L oxygen sample contains
\[
N_{\mathrm{O_2}}=N
\]
Therefore, the total number of initial reactant molecules is
\[
N_{\text{reactants}}=2N+N=3N
\]
\(\textbf{Product molecular count:}\)
The volume of water vapour produced is 2 L under the same temperature-pressure conditions.
Therefore,
\[
N_{\text{products}}=2N
\]
\(\textbf{Required ratio:}\)
\[
\frac{N_{\text{products}}}{N_{\text{reactants}}}=\frac{2N}{3N}
\]
\[
\frac{N_{\text{products}}}{N_{\text{reactants}}}=\frac{2}{3}
\]
\(\textbf{Chemical consistency:}\)
Although the total number of gas molecules decreases, the numbers of hydrogen and oxygen atoms remain conserved.
Distinguishing atoms from diatomic molecules is essential for explaining this gas-volume relationship.
Dalton's original atomic treatment alone could not adequately explain the observed laws of gaseous volumes.
\(\textbf{Final answer:}\)
\[
\boxed{\frac{2}{3}}
\]
16. Two identical reactions are carried out using 9 g of solid A and 21 g of solid B. In each reaction, the reactants are completely consumed and the products consist of 24 g of solid residue and a gas. In Trial I, the reaction occurs in a sealed vessel, and its total contents retain a mass of 30 g. In Trial II, the vessel is open and the gas escapes completely, leaving 24 g of solid residue. No material other than the gaseous product enters or leaves either system. Which conclusion correctly reconciles the observations with Dalton's atomic theory?
ⓐ. A 6 g gaseous product forms in both trials; its escape reduces the total mass of all reaction products.
ⓑ. A 6 g gaseous product forms only in the sealed trial; the open reaction produces less total matter.
ⓒ. A 6 g portion of matter is destroyed in both trials; sealing merely prevents detection of this loss.
ⓓ. A 6 g gaseous product forms in both trials; its escape reduces the mass retained in the open vessel.
Correct Answer: A 6 g gaseous product forms in both trials; its escape reduces the mass retained in the open vessel.
Explanation: \(\textbf{Initial mass of reactants:}\)
The combined starting mass in each trial is
\[
m_{\text{initial}}=9+21
\]
\[
m_{\text{initial}}=30\,\text{g}
\]
\(\textbf{Applying atom and mass conservation:}\)
According to Dalton's atomic theory, chemical reactions rearrange atoms rather than creating or destroying them.
Therefore, the total mass of products must equal the total mass of reactants when all products are accounted for.
The solid product has a mass of 24 g.
The remaining product mass must belong to the gas.
\[
m_{\text{gas}}=30-24
\]
\[
m_{\text{gas}}=6\,\text{g}
\]
\(\textbf{Interpretation of Trial I:}\)
In the sealed vessel, both the solid residue and the gaseous product remain inside.
Therefore,
\[
m_{\text{contents}}=24+6
\]
\[
m_{\text{contents}}=30\,\text{g}
\]
The measured result agrees with conservation of total mass.
\(\textbf{Interpretation of Trial II:}\)
In the open vessel, the 6 g gaseous product escapes into the surroundings.
The measured 24 g represents only the retained solid residue, not all the reaction products.
If the escaped gas is included in the mass accounting, the product total is again 30 g.
\(\textbf{Critical distinction:}\)
A decrease in the mass remaining inside an open vessel does not imply that atoms have been destroyed.
Mass conservation applies to the complete reacting system, including material that has escaped the measurement boundary.
Both trials are therefore consistent with Dalton's atom-rearrangement principle.
\(\textbf{Final conclusion:}\)
Both reactions produce 6 g of gas, but only the open vessel loses that gas to the surroundings.
17. Two pure ionic compounds, \(\mathrm{Al_2(SO_4)_3}\) and \(\mathrm{Na_2SO_4}\), are taken in quantities containing equal numbers of sulphate ions. What are the respective ratios of aluminium atoms to sodium atoms and oxygen atoms in the first sample to oxygen atoms in the second sample?
ⓐ. \(1:3\) and \(1:1\)
ⓑ. \(3:1\) and \(1:1\)
ⓒ. \(1:3\) and \(3:1\)
ⓓ. \(1:1\) and \(1:3\)
Correct Answer: \(1:3\) and \(1:1\)
Explanation: \(\textbf{Constituent-ion interpretation:}\)
Both compounds are ionic substances, so their chemical formulae specify formula-unit compositions rather than discrete molecular structures.
One formula unit of \(\mathrm{Al_2(SO_4)_3}\) contains two aluminium ions and three sulphate ions.
One formula unit of \(\mathrm{Na_2SO_4}\) contains two sodium ions and one sulphate ion.
\(\textbf{Equal sulphate-ion condition:}\)
Let the common number of sulphate ions in each sample be \(N\).
The numbers of formula units required are
\[
N_{\mathrm{Al_2(SO_4)_3}}=\frac{N}{3}
\]
\[
N_{\mathrm{Na_2SO_4}}=N
\]
\(\textbf{Comparison of metal atoms:}\)
Each aluminium sulphate formula unit contains two aluminium atoms.
Therefore,
\[
N_{\mathrm{Al}}=2\times\frac{N}{3}=\frac{2N}{3}
\]
Each sodium sulphate formula unit contains two sodium atoms.
Therefore,
\[
N_{\mathrm{Na}}=2N
\]
The required metal-atom ratio is
\[
N_{\mathrm{Al}}:N_{\mathrm{Na}}=\frac{2N}{3}:2N
\]
\[
N_{\mathrm{Al}}:N_{\mathrm{Na}}=1:3
\]
\(\textbf{Comparison of oxygen atoms:}\)
Every sulphate ion contains four oxygen atoms.
Since both samples contain \(N\) sulphate ions,
\[
N_{\mathrm{O,first}}=4N
\]
\[
N_{\mathrm{O,second}}=4N
\]
Hence,
\[
N_{\mathrm{O,first}}:N_{\mathrm{O,second}}=1:1
\]
The two oxygen-atom counts are equal even though the numbers of metal atoms differ.
The essential distinction is between formula units, constituent polyatomic ions and individual atoms.
\(\textbf{Final answer:}\)
\[
\boxed{(1:3,\ 1:1)}
\]
18. A gaseous mixture contains only two kinds of molecules, \(\mathrm{X_2Y}\) and \(\mathrm{XY_2}\), where X and Y represent different elements. The mixture contains a total of 23 atoms of X and 19 atoms of Y. What fraction of the molecules present is \(\mathrm{X_2Y}\)?
ⓐ. \(\dfrac{5}{14}\)
ⓑ. \(\dfrac{9}{23}\)
ⓒ. \(\dfrac{9}{14}\)
ⓓ. \(\dfrac{14}{23}\)
Correct Answer: \(\dfrac{9}{14}\)
Explanation: \(\textbf{Identifying the unknown molecular populations:}\)
Let the number of \(\mathrm{X_2Y}\) molecules be \(a\).
Let the number of \(\mathrm{XY_2}\) molecules be \(b\).
Each \(\mathrm{X_2Y}\) molecule contributes two X atoms and one Y atom.
Each \(\mathrm{XY_2}\) molecule contributes one X atom and two Y atoms.
\(\textbf{Using the X-atom count:}\)
The total number of X atoms is 23.
Therefore,
\[
2a+b=23 \qquad (1)
\]
\(\textbf{Using the Y-atom count:}\)
The total number of Y atoms is 19.
Therefore,
\[
a+2b=19 \qquad (2)
\]
\(\textbf{Solving the simultaneous constraints:}\)
From Equation (1),
\[
b=23-2a
\]
Substituting into Equation (2),
\[
a+2(23-2a)=19
\]
\[
a+46-4a=19
\]
\[
3a=27
\]
\[
a=9
\]
Consequently,
\[
b=23-2(9)=5
\]
\(\textbf{Determining the molecular fraction:}\)
The total number of molecules is
\[
N_{\mathrm{total}}=9+5=14
\]
The required fraction is
\[
f_{\mathrm{X_2Y}}=\frac{9}{14}
\]
\(\textbf{Independent atom-count check:}\)
The calculated numbers must reproduce both given elemental counts.
\[
N_X=2(9)+5=23
\]
\[
N_Y=9+2(5)=19
\]
Both constraints are satisfied.
The molecular fraction differs from the fraction of all constituent atoms contributed by either molecular species.
\(\textbf{Final answer:}\)
\[
\boxed{\frac{9}{14}}
\]
19. A sealed vessel initially contains exactly 15 molecules of \(\mathrm{O_2}\). Under suitable conditions, all these molecules are converted into exactly 10 molecules of \(\mathrm{O_3}\), with no other chemical species present. Which statement correctly describes the change?
ⓐ. The molecular count is conserved, but the oxygen-atom count decreases.
ⓑ. The molecular count decreases, while the elemental identity remains unchanged.
ⓒ. A compound is formed because every product molecule contains three atoms.
ⓓ. A different chemical element is formed because the molecular formula changes.
Correct Answer: The molecular count decreases, while the elemental identity remains unchanged.
Explanation: Oxygen can exist as different molecular forms, including dioxygen and ozone. A dioxygen molecule, \(\mathrm{O_2}\), contains two oxygen atoms, whereas an ozone molecule, \(\mathrm{O_3}\), contains three oxygen atoms. Both are elemental substances because their molecules contain atoms of only oxygen. Initially, 15 dioxygen molecules contain \(15\times2=30\) oxygen atoms. Finally, 10 ozone molecules contain \(10\times3=30\) oxygen atoms. The total number of oxygen atoms therefore remains unchanged. However, the number of molecules decreases from 15 to 10. This demonstrates that conservation of atoms does not require conservation of molecular count. Conversion between these molecular forms changes chemical bonding and molecular identity without producing a different chemical element. The product is not a compound merely because each molecule contains three atoms. A compound must contain chemically combined atoms of different elements. The fact that both molecular forms are composed exclusively of oxygen is decisive for their elemental classification. Thus, the change preserves elemental identity and atom count but not molecular count.
20. A hypothetical element E has two isotopes of masses \(10\,\mathrm{u}\) and \(11\,\mathrm{u}\). Sample P contains \(75\%\) of the lighter isotope and \(25\%\) of the heavier isotope by number of atoms. Sample Q contains \(25\%\) of the lighter isotope and \(75\%\) of the heavier isotope by number of atoms. Which conclusion is correct?
ⓐ. P and Q are different elements, and their average atomic masses differ by \(0.50\,\mathrm{u}\).
ⓑ. P and Q are the same element, and their average atomic masses differ by \(1.00\,\mathrm{u}\).
ⓒ. P and Q are different compounds, and their average atomic masses differ by \(0.25\,\mathrm{u}\).
ⓓ. P and Q are samples of the same element, and their average atomic masses differ by \(0.50\,\mathrm{u}\).
Correct Answer: P and Q are samples of the same element, and their average atomic masses differ by \(0.50\,\mathrm{u}\).
Explanation: \(\textbf{Isotopic identity:}\)
Isotopes of an element have the same atomic number but different masses.
Different relative abundances of its isotopes can therefore produce different average atomic masses.
The two samples still contain atoms of the same chemical element.
\(\textbf{Average atomic mass of sample P:}\)
For isotopic abundances expressed as fractions by number,
\[
\overline{M}=\sum_i f_iM_i
\]
where \(f_i\) is the fractional abundance of isotope \(i\).
For sample P,
\[
\overline{M}_P=(0.75)(10)+(0.25)(11)
\]
\[
\overline{M}_P=7.50+2.75
\]
\[
\overline{M}_P=10.25\,\mathrm{u}
\]
\(\textbf{Average atomic mass of sample Q:}\)
\[
\overline{M}_Q=(0.25)(10)+(0.75)(11)
\]
\[
\overline{M}_Q=2.50+8.25
\]
\[
\overline{M}_Q=10.75\,\mathrm{u}
\]
\(\textbf{Comparison:}\)
The difference between the average atomic masses is
\[
\Delta\overline{M}=10.75-10.25
\]
\[
\Delta\overline{M}=0.50\,\mathrm{u}
\]
\(\textbf{Dalton's historical postulate:}\)
Dalton originally proposed that atoms of the same element possess identical masses.
The existence of isotopes requires modification of that original postulate.
A change in isotopic abundance does not automatically imply a change in chemical elemental identity.
Both P and Q remain elemental samples, even though their average atomic masses differ.
\(\textbf{Final answer:}\)
\[
\boxed{0.50\,\mathrm{u}}
\]
