201. In a closed vessel, \(4.8\,g\) of magnesium reacts completely with oxygen to form \(8.0\,g\) of magnesium oxide. What mass of oxygen has combined with magnesium?
ⓐ. \(1.6\,g\)
ⓑ. \(4.8\,g\)
ⓒ. \(12.8\,g\)
ⓓ. \(3.2\,g\)
Correct Answer: \(3.2\,g\)
Explanation: \( \textbf{Given data:} \) mass of magnesium \(=4.8\,g\), mass of magnesium oxide formed \(=8.0\,g\).
\( \textbf{System condition:} \) the reaction occurs in a closed vessel.
\( \textbf{Law used:} \) conservation of mass.
\( \textbf{Meaning:} \) total mass of reactants \(=\) total mass of products.
\( \textbf{Reactants:} \) magnesium \(+\) oxygen.
\[
m_{\mathrm{Mg}}+m_{\mathrm{O}}=m_{\mathrm{MgO}}
\]
\( \textbf{Substitution:} \)
\[
4.8\,g+m_{\mathrm{O}}=8.0\,g
\]
\( \textbf{Solve for oxygen:} \)
\[
m_{\mathrm{O}}=8.0\,g-4.8\,g=3.2\,g
\]
\( \textbf{Final answer:} \) \(3.2\,g\) of oxygen combines with magnesium.
The product mass is not added again; it already contains the mass of both reactants.
202. A reaction is performed in four closed containers. The recorded masses are shown below.
| Container | Total mass before reaction | Total mass after reaction |
|---|
| P | \(18.5\,g\) | \(18.5\,g\) |
| Q | \(22.0\,g\) | \(21.2\,g\) |
| R | \(40.3\,g\) | \(40.3\,g\) |
| S | \(9.75\,g\) | \(9.75\,g\) |
The record that needs rechecking for a closed-system reaction is:
ⓐ. Container P
ⓑ. Container Q
ⓒ. Container R
ⓓ. Container S
Correct Answer: Container Q
Explanation: In a closed system, no matter is allowed to enter or escape during the reaction. The law of conservation of mass therefore requires the total mass before reaction to equal the total mass after reaction. Containers P, R, and S show equal before-and-after masses. Container Q shows a decrease from \(22.0\,g\) to \(21.2\,g\), which is not expected for a properly closed system. Such a record suggests gas escape, transfer loss, leakage, or measurement error rather than failure of the law.
203. A metal carbonate is heated in a test tube fitted with a delivery tube leading to a gas jar. If only the test tube is weighed after heating, the measured mass is lower. The best reason is:
ⓐ. The reaction destroys part of the mass
ⓑ. The collected gas is outside the weighed test tube
ⓒ. Solids have mass but gases do not
ⓓ. Heating changes part of the measured mass into temperature
Correct Answer: The collected gas is outside the weighed test tube
Explanation: When a carbonate decomposes, a gaseous product may leave the heated tube. If the gas is collected elsewhere, the test tube alone no longer contains all products of the reaction. Conservation of mass applies to the entire system, including the solid residue, the gas, and any connected apparatus. Weighing only one part of the system can make mass appear to decrease. The boundary chosen for measurement must include all matter produced or remaining.
204. In the reaction record below, calcium carbonate decomposes in a sealed apparatus.
Mass of sealed apparatus before heating \(=126.50\,g\). Mass of sealed apparatus after complete cooling \(=126.50\,g\). Inside the apparatus, solid calcium oxide and carbon dioxide gas are present after heating.
The observation mainly supports:
ⓐ. Closed-system mass conservation
ⓑ. Definite proportions in a mixture
ⓒ. Loss of mass during gas formation
ⓓ. Conversion of gas into energy
Correct Answer: Closed-system mass conservation
Explanation: The sealed apparatus represents a closed system because matter cannot escape. Even though calcium carbonate changes into calcium oxide and carbon dioxide, the total mass of the sealed apparatus remains the same. The gas formed is still inside the apparatus and is included in the mass measurement. This directly supports conservation of mass. The observation is about total mass before and after reaction, not about the fixed composition of one compound.
205. A student balances the equation for hydrogen and oxygen forming water as \(2\mathrm{H_2}+\mathrm{O_2}\rightarrow2\mathrm{H_2O}\). The balancing is connected with conservation of mass because:
ⓐ. Atoms of each element are equal on both sides
ⓑ. It gives the same number of molecules on both sides
ⓒ. It changes oxygen atoms into hydrogen atoms before reaction
ⓓ. It removes the need to know any chemical formulae
Correct Answer: Atoms of each element are equal on both sides
Explanation: A balanced equation conserves atoms of every element. In \(2\mathrm{H_2}+\mathrm{O_2}\rightarrow2\mathrm{H_2O}\), there are \(4\) hydrogen atoms and \(2\) oxygen atoms on each side. Equal atom counts mean the same kinds and numbers of atoms are present before and after the reaction. Since atoms have mass, conserving atoms is consistent with conserving total mass. The number of molecules need not be the same on both sides; here \(3\) reactant molecules form \(2\) product molecules.
206. A compound is prepared from copper and oxygen. In one sample, \(3.18\,g\) of copper combines with \(0.80\,g\) of oxygen. In another sample of the same compound, \(6.36\,g\) of copper combines with \(1.60\,g\) of oxygen. This illustrates:
ⓐ. Law of multiple proportions
ⓑ. Avogadro's law
ⓒ. Law of definite proportions
ⓓ. Gay-Lussac's law of gaseous volumes
Correct Answer: Law of definite proportions
Explanation: The two samples contain copper and oxygen in the same mass ratio. In the first sample, the ratio of copper to oxygen is \(3.18:0.80\). In the second sample, both masses are doubled, giving \(6.36:1.60\). The relative composition remains unchanged even though the sample size changes. This is the idea of definite proportions: a given compound always contains the same elements in the same fixed mass ratio.
207. Pure water samples obtained after purification from different sources always contains hydrogen and oxygen in the same mass ratio. This statement is an example of:
ⓐ. Law of conservation of mass
ⓑ. Law of definite proportions
ⓒ. Law of multiple proportions
ⓓ. Dalton's law of partial pressure
Correct Answer: Law of definite proportions
Explanation: The law of definite proportions says that a given compound has a fixed composition by mass. Pure water is the same compound, \(\mathrm{H_2O}\), regardless of its source. Therefore, its hydrogen-to-oxygen mass ratio remains constant. The source of the sample does not change the composition if the sample is chemically pure water. This law separates a pure compound from mixtures, where composition may vary.
208. A mixture of salt and sand can contain \(20\%\) salt in one sample and \(60\%\) salt in another sample. This does not obey the fixed-ratio behaviour of compounds because:
ⓐ. Mixtures have variable composition
ⓑ. Mixtures always contain only one element
ⓒ. Mixtures have no mass
ⓓ. Mixtures are always gases
Correct Answer: Mixtures have variable composition
Explanation: A mixture is a physical combination of substances, so its components can be present in different proportions. Salt and sand can be mixed in many possible mass ratios without forming a new compound. A compound, in contrast, has chemically combined elements in a fixed ratio. The law of definite proportions applies to pure compounds, not to ordinary mixtures. Variable composition is one of the main clues that a sample is a mixture.
209. A compound contains \(2.0\,g\) of hydrogen for every \(16.0\,g\) of oxygen. If a pure sample of the same compound contains \(64.0\,g\) of oxygen, what mass of hydrogen should it contain?
ⓐ. \(2.0\,g\)
ⓑ. \(4.0\,g\)
ⓒ. \(8.0\,g\)
ⓓ. \(32.0\,g\)
Correct Answer: \(8.0\,g\)
Explanation: \( \textbf{Given fixed ratio:} \) hydrogen : oxygen \(=2.0\,g:16.0\,g\).
\( \textbf{Simplify the ratio:} \)
\[
\frac{2.0}{16.0}=\frac{1}{8}
\]
\( \textbf{Meaning:} \) hydrogen mass is \(\frac{1}{8}\) of the oxygen mass in this compound.
\( \textbf{Given oxygen mass:} \) \(64.0\,g\).
\( \textbf{Find hydrogen mass:} \)
\[
m_{\mathrm{H}}=\frac{1}{8}\times64.0\,g
\]
\( \textbf{Calculation:} \)
\[
m_{\mathrm{H}}=8.0\,g
\]
\( \textbf{Final answer:} \) the sample contains \(8.0\,g\) of hydrogen.
The ratio is scaled up with sample size, but the composition of the compound remains fixed.
210. A data table for a compound made of elements X and Y is shown below.
| Sample | Mass of X | Mass of Y |
|---|
| P | \(5.0\,g\) | \(2.0\,g\) |
| Q | \(10.0\,g\) | \(4.0\,g\) |
| R | \(15.0\,g\) | \(6.0\,g\) |
| S | \(20.0\,g\) | \(9.0\,g\) |
The sample that is not consistent with the same compound is:
ⓐ. Sample P
ⓑ. Sample Q
ⓒ. Sample R
ⓓ. Sample S
Correct Answer: Sample S
Explanation: \( \textbf{Check mass ratio for P:} \)
\[
X:Y=5.0:2.0=2.5:1
\]
\( \textbf{Check mass ratio for Q:} \)
\[
X:Y=10.0:4.0=2.5:1
\]
\( \textbf{Check mass ratio for R:} \)
\[
X:Y=15.0:6.0=2.5:1
\]
\( \textbf{Check mass ratio for S:} \)
\[
X:Y=20.0:9.0\approx2.22:1
\]
\( \textbf{Comparison:} \) P, Q, and R have the same mass ratio, while S differs.
\( \textbf{Final answer:} \) Sample S is not consistent with the same compound.
For one pure compound, changing sample size should not change the mass ratio of its elements.
211. Assertion: Pure carbon dioxide from dry ice and pure carbon dioxide from burning carbon have the same percentage composition.
Reason: A given compound always contains the same elements in the same fixed mass ratio.
ⓐ. Both Assertion and Reason are true, and Reason explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: The assertion is true because pure carbon dioxide is the same compound, \(\mathrm{CO_2}\), regardless of how it is obtained. Its carbon and oxygen are chemically combined in a fixed ratio. The reason states the law of definite proportions, which directly explains the assertion. Different sources do not change the composition if the compound is pure. Source-based variation is possible for mixtures, but not for a pure compound with a definite formula.
212. A sample of a compound contains \(40.0\%\) carbon and \(60.0\%\) oxygen by mass. Another pure sample of the same compound has mass \(25.0\,g\). What mass of oxygen is present?
ⓐ. \(10.0\,g\)
ⓑ. \(20.0\,g\)
ⓒ. \(60.0\,g\)
ⓓ. \(15.0\,g\)
Correct Answer: \(15.0\,g\)
Explanation: \( \textbf{Given composition:} \) oxygen \(=60.0\%\) by mass.
\( \textbf{Sample mass:} \) \(25.0\,g\).
\( \textbf{Required quantity:} \) mass of oxygen in the sample.
\( \textbf{Percentage relation:} \)
\[
\text{mass of oxygen}=\frac{60.0}{100}\times\text{mass of compound}
\]
\( \textbf{Substitution:} \)
\[
m_{\mathrm{O}}=\frac{60.0}{100}\times25.0\,g
\]
\( \textbf{Calculation:} \)
\[
m_{\mathrm{O}}=15.0\,g
\]
\( \textbf{Final answer:} \) \(15.0\,g\) of oxygen is present.
The same percentage composition is used because the sample is stated to be the same pure compound.
213. A sample contains \(9\,g\) of water mixed with \(1\,g\) of salt. Another sample contains \(9\,g\) of water mixed with \(3\,g\) of salt. These two samples mainly show:
ⓐ. A mixture can have variable composition
ⓑ. Water has no fixed composition
ⓒ. Salt and water must form a new compound in every ratio
ⓓ. Definite proportions apply to all mixtures
Correct Answer: A mixture can have variable composition
Explanation: Salt water is a mixture, not a pure compound with one fixed formula. The amount of salt dissolved in water can vary from one sample to another. Both samples contain water and salt, but their component proportions are different. This variation is normal for mixtures. The fixed-ratio idea belongs to a given compound, not to a solution prepared with different amounts of solute.
214. Two pure samples of sodium chloride are analysed.
| Sample | Mass of sodium | Mass of chlorine |
|---|
| P | \(2.30\,g\) | \(3.55\,g\) |
| Q | \(4.60\,g\) | \(7.10\,g\) |
The best interpretation is:
ⓐ. The data show a mixture because the second sample has larger masses
ⓑ. The data reject fixed composition because both elements have different masses in Q
ⓒ. The data show conservation of mass only if gas escapes
ⓓ. Same sodium-to-chlorine mass ratio supports definite proportions
Correct Answer: Same sodium-to-chlorine mass ratio supports definite proportions
Explanation: \( \textbf{Sample P ratio:} \)
\[
\frac{2.30\,g}{3.55\,g}\approx0.648
\]
\( \textbf{Sample Q ratio:} \)
\[
\frac{4.60\,g}{7.10\,g}\approx0.648
\]
\( \textbf{Comparison:} \) the sodium-to-chlorine ratio is the same in both samples.
\( \textbf{Interpretation:} \) larger sample mass changes the actual masses, not the mass ratio.
\( \textbf{Law involved:} \) this supports the law of definite proportions.
\( \textbf{Final answer:} \) both samples are consistent with pure sodium chloride.
A larger sample of the same compound should contain proportionally larger masses of each element.
215. A learner says, "If two samples have different total masses, they cannot be the same compound." The best reply is:
ⓐ. Same compound: different mass, same composition ratio
ⓑ. A compound exists only in one fixed laboratory sample mass
ⓒ. Different total mass always means a different compound identity
ⓓ. Compounds can be identified by total mass without composition
Correct Answer: Same compound: different mass, same composition ratio
Explanation: A compound has fixed composition, not a fixed sample size. A small sample and a large sample of pure water are both \(\mathrm{H_2O}\). Their total masses differ, but the ratio of hydrogen to oxygen by mass remains the same. Chemical identity depends on composition and structure, not on how much sample is taken. This is why mass ratios are compared instead of total sample masses when applying the law of definite proportions.
216. For a pure compound containing elements A and B, a graph is plotted with mass of B on the y-axis and mass of A on the x-axis for different sample sizes. The graph is a straight line through the origin. The slope represents:
ⓐ. The total mass of every sample
ⓑ. The number of physical states present
ⓒ. The gas volume formed per reaction
ⓓ. The fixed mass ratio of B to A
Correct Answer: The fixed mass ratio of B to A
Explanation: \( \textbf{Graph axes:} \) y-axis \(=\) mass of B, x-axis \(=\) mass of A.
\( \textbf{Slope meaning:} \)
\[
\text{slope}=\frac{\Delta(\text{mass of B})}{\Delta(\text{mass of A})}
\]
\( \textbf{For one pure compound:} \) the mass ratio of elements remains fixed.
\( \textbf{Straight-line meaning:} \) as sample size increases, masses of A and B increase proportionally.
\( \textbf{Origin meaning:} \) zero mass of compound would contain zero mass of both elements.
\( \textbf{Final answer:} \) the slope gives the fixed mass ratio \(\frac{\text{mass of B}}{\text{mass of A}}\).
This graph form is a visual way of expressing the law of definite proportions.
217. The law of multiple proportions is most directly applied when:
ⓐ. Two elements form more than one compound and one element mass is fixed
ⓑ. A compound is prepared from different sources but has the same composition
ⓒ. Total mass before and after reaction is compared in a closed container
ⓓ. Equal gas volumes are compared at different temperatures and pressures
Correct Answer: Two elements form more than one compound and one element mass is fixed
Explanation: The law of multiple proportions applies when the same two elements form two or more different compounds. A fixed mass of one element is chosen, and the masses of the other element that combine with it are compared. These masses should be in a simple whole-number ratio. The law of definite proportions deals with the fixed composition of one compound, not comparison of several compounds. The phrase "more than one compound" is essential because the law compares different combining ratios of the same elements.
218. Carbon and oxygen form \(\mathrm{CO}\) and \(\mathrm{CO_2}\). For the same mass of carbon, the masses of oxygen that combine are in the ratio:
ⓐ. \(1:1\)
ⓑ. \(1:2\)
ⓒ. \(2:1\)
ⓓ. \(3:2\)
Correct Answer: \(1:2\)
Explanation: In \(\mathrm{CO}\), one carbon atom is combined with one oxygen atom. In \(\mathrm{CO_2}\), the same one carbon atom is combined with two oxygen atoms. If the mass of carbon is fixed, the oxygen mass in \(\mathrm{CO_2}\) is double the oxygen mass in \(\mathrm{CO}\). Therefore, the oxygen masses are in the ratio \(1:2\). This is a standard illustration of the law of multiple proportions because the same elements form two different compounds.
219. A data set for two compounds of nitrogen and oxygen is given below.
| Compound | Mass of nitrogen | Mass of oxygen |
|---|
| P | \(14\,g\) | \(16\,g\) |
| Q | \(14\,g\) | \(32\,g\) |
The ratio of oxygen masses combining with the fixed mass of nitrogen is:
ⓐ. \(1:1\)
ⓑ. \(1:2\)
ⓒ. \(2:3\)
ⓓ. \(7:8\)
Correct Answer: \(1:2\)
Explanation: \( \textbf{Fixed element:} \) nitrogen has the same mass, \(14\,g\), in both compounds.
\( \textbf{Compare oxygen masses:} \) compound P has \(16\,g\) oxygen and compound Q has \(32\,g\) oxygen.
\( \textbf{Set up the ratio:} \)
\[
16:32
\]
\( \textbf{Simplify:} \)
\[
16:32=1:2
\]
\( \textbf{Law connection:} \) for a fixed mass of nitrogen, oxygen combines in a simple whole-number ratio.
\( \textbf{Final answer:} \) the oxygen mass ratio is \(1:2\).
The nitrogen masses should not be compared again because they were already fixed by the data.
220. A learner compares \(\mathrm{CO}\) and \(\mathrm{CO_2}\) by fixing oxygen mass instead of carbon mass. To apply the law of multiple proportions in the usual way for these formulae, the better comparison is:
ⓐ. Fix the mass of carbon and compare the masses of oxygen
ⓑ. Fix the total mass of both compounds and compare colours
ⓒ. Fix the volume of the container and compare densities only
ⓓ. Compare the melting points of carbon and oxygen separately
Correct Answer: Fix the mass of carbon and compare the masses of oxygen
Explanation: The law of multiple proportions requires a fixed mass of one element. For \(\mathrm{CO}\) and \(\mathrm{CO_2}\), fixing carbon is convenient because each formula contains one carbon atom. Then the oxygen amounts are directly compared as one oxygen atom in \(\mathrm{CO}\) and two oxygen atoms in \(\mathrm{CO_2}\). Comparing total compound mass or physical properties does not test the law. The ratio must be based on combining masses of elements, not on appearance or container conditions.
