1. A true solution is best described as a mixture in which the components are:
ⓐ. present in equal masses throughout the sample
ⓑ. uniformly distributed throughout the mixture
ⓒ. chemically combined to form a new pure substance
ⓓ. separated into visible layers after standing
Correct Answer: uniformly distributed throughout the mixture
Explanation: A true solution is a homogeneous mixture, so its composition remains uniform throughout the sample. The solute particles are distributed at molecular or ionic level in the solvent. Equal masses of components are not required; a small amount of salt can form a true solution in a much larger amount of water. Chemical combination is also not required because solution formation is usually physical mixing. Visible layers indicate non-uniformity, so they do not fit the idea of a true solution.
2. Salt water is treated as a solution rather than a new compound mainly because:
ⓐ. salt and water are chemically combined in a fixed ratio
ⓑ. water is consumed completely when the salt dissolves
ⓒ. salt is converted permanently into a different substance
ⓓ. salt and water stay uniformly mixed without reacting
Correct Answer: salt and water stay uniformly mixed without reacting
Explanation: In salt water, the ions from salt spread uniformly through water, but the components do not combine to form a new pure compound. The composition of a solution can vary, so salt water need not have one fixed mass ratio. Water does not disappear; it acts as the solvent medium. The dissolved salt can often be recovered by physical methods such as evaporation of water. The key idea is homogeneous physical mixing, not permanent chemical conversion.
3. In ordinary solute-solvent language, the solute is usually the component that:
ⓐ. is usually present in smaller amount and is dissolved
ⓑ. usually determines the physical state of the solution
ⓒ. must be a solid before the solution is prepared
ⓓ. must be water whenever the final solution is liquid
Correct Answer: is usually present in smaller amount and is dissolved
Explanation: The solute is commonly the component present in the smaller amount and dissolved in another component. The solvent is usually present in the larger amount and provides the dissolving medium. This smaller-amount rule is a useful starting point, but it should not be treated as an absolute definition in every special case. A solute may be solid, liquid, or gas depending on the solution. Water is a common solvent, but solution chemistry is not limited to water-based mixtures.
4. A data sheet lists some basic quantities used while discussing solutions.
| Quantity | Common unit symbol |
|---|
| P. Mass | \(g\) |
| Q. Volume | \(mL\) |
| R. Amount of substance | \(mol\) |
| S. Temperature | \(K\) |
The entry that connects solution composition directly with the mole concept is:
ⓐ. P only
ⓑ. R only
ⓒ. P and Q only
ⓓ. Q and S only
Correct Answer: R only
Explanation: The mole concept is directly represented by the amount of substance, whose common unit is \(mol\). Mass in \(g\) can be converted into moles only when molar mass is known, so it is not itself the mole quantity. Volume in \(mL\) is useful for preparing and measuring solutions but does not directly count particles. Temperature in \(K\) is important in some solution laws, especially gas-related relations, but it is not the amount of substance. The symbol \(mol\) is the direct link between composition and particle amount.
5. The missing conversion in the statement “\(1\,L=\) ______ \(mL\)” is:
ⓐ. \(10\,mL\)
ⓑ. \(100\,mL\)
ⓒ. \(1000\,mL\)
ⓓ. \(10^6\,mL\)
Correct Answer: \(1000\,mL\)
Explanation: \( \textbf{Required conversion:} \) Convert litre into millilitre.
\( \textbf{Prefix meaning:} \) The prefix milli means \(10^{-3}\) of the base unit.
\( \textbf{Relation:} \) \(1\,mL=10^{-3}\,L\).
\( \textbf{Rearrangement:} \) If \(1\,mL=0.001\,L\), then \(1\,L=1000\,mL\).
\( \textbf{Use in solutions:} \) This conversion is needed when volume is given in \(mL\) but a concentration expression requires volume in \(L\).
\( \textbf{Final answer:} \) The missing value is \(1000\,mL\), not \(100\,mL\), because three powers of \(10\) separate \(L\) and \(mL\).
6. The terms dilute and concentrated mainly compare:
ⓐ. solute amount relative to a stated composition basis
ⓑ. solution colour relative to a stated visual reference
ⓒ. reaction rate relative to the amount of solvent present
ⓓ. component boiling points relative to room temperature
Correct Answer: solute amount relative to a stated composition basis
Explanation: Dilute and concentrated are comparative composition terms. A dilute solution contains relatively less solute, while a concentrated solution contains relatively more solute for a stated basis of comparison. These words do not automatically give an exact numerical concentration unless the basis and data are supplied. Colour can sometimes give a rough visual clue for coloured solutions, but it is not the definition. The comparison must be tied to amount of solute relative to solution or solvent, otherwise the words can be misleading.
7. A substance has molar mass \(40\,g\,mol^{-1}\). The amount of substance in \(20\,g\) of it is:
ⓐ. \(0.25\,mol\)
ⓑ. \(800\,mol\)
ⓒ. \(2.00\,mol\)
ⓓ. \(0.50\,mol\)
Correct Answer: \(0.50\,mol\)
Explanation: \( \textbf{Known data:} \) Mass of substance \(=20\,g\), molar mass \(=40\,g\,mol^{-1}\).
\( \textbf{Required quantity:} \) Amount of substance in \(mol\).
\( \textbf{Mole relation:} \)
\[
n=\frac{\text{given mass}}{\text{molar mass}}
\]
\( \textbf{Substitution:} \)
\[
n=\frac{20\,g}{40\,g\,mol^{-1}}
\]
\( \textbf{Cancellation of unit:} \) The \(g\) unit cancels, leaving \(mol\).
\( \textbf{Calculation:} \)
\[
n=0.50\,mol
\]
\( \textbf{Final answer:} \) The amount is \(0.50\,mol\). Using \(40 \div 20\) would invert the required relation and give a value for the wrong comparison.
8. A solvent mass is recorded as \(0.250\,kg\). The same mass in grams is:
ⓐ. \(0.250\,g\)
ⓑ. \(25.0\,g\)
ⓒ. \(250\,g\)
ⓓ. \(2500\,g\)
Correct Answer: \(250\,g\)
Explanation: \( \textbf{Known relation:} \) \(1\,kg=1000\,g\).
\( \textbf{Given mass:} \) \(0.250\,kg\).
\( \textbf{Conversion step:} \)
\[
0.250\,kg \times \frac{1000\,g}{1\,kg}
\]
\( \textbf{Unit cancellation:} \) The \(kg\) unit cancels and the answer is expressed in \(g\).
\( \textbf{Calculation:} \)
\[
0.250\times1000=250
\]
\( \textbf{Final answer:} \) The mass is \(250\,g\). This conversion is especially useful before deciding whether a formula needs solvent mass in \(kg\) or a mass recorded in \(g\).
9. A gas-related solution relation requires temperature in \(K\). A temperature of \(27^{\circ}C\) should be entered as:
ⓐ. \(300\,K\)
ⓑ. \(273\,K\)
ⓒ. \(246\,K\)
ⓓ. \(27\,K\)
Correct Answer: \(300\,K\)
Explanation: \( \textbf{Temperature scale needed:} \) Gas-related laws commonly use absolute temperature in \(K\).
\( \textbf{Conversion relation:} \)
\[
T(K)=T(^{\circ}C)+273
\]
\( \textbf{Substitution:} \)
\[
T=27+273
\]
\( \textbf{Calculation:} \)
\[
T=300\,K
\]
\( \textbf{Final answer:} \) The required temperature is \(300\,K\). Writing \(27\,K\) keeps the Celsius number but changes the physical scale, which would make later gas-law calculations inconsistent.
10. Case P contains \(5\,g\) of sugar in \(100\,g\) of water, while Case Q contains \(10\,g\) of sugar in \(1000\,g\) of water. Based on solute-to-solvent mass ratio:
ⓐ. Case Q, because it contains more sugar in total
ⓑ. Both cases, because their sugar masses differ proportionally
ⓒ. Case P, because its solute-to-solvent mass ratio is larger
ⓓ. Case Q, because its total solution mass is larger
Correct Answer: Case P, because its solute-to-solvent mass ratio is larger
Explanation: \( \textbf{Comparison basis:} \) Use solute mass relative to solvent mass.
\( \textbf{Case P ratio:} \)
\[
\frac{5\,g}{100\,g}=0.050
\]
\( \textbf{Case Q ratio:} \)
\[
\frac{10\,g}{1000\,g}=0.010
\]
\( \textbf{Interpretation:} \) \(0.050\) is greater than \(0.010\), so Case P has more sugar per unit mass of water.
\( \textbf{Why total sugar alone is insufficient:} \) Case Q has more sugar in total, but it also has much more water.
\( \textbf{Final answer:} \) Case P is more concentrated on the stated mass-ratio basis. The comparison changes from absolute amount to relative amount once concentration is discussed.
11. Use the descriptions below. Vessel P contains sugar dissolved completely in water, Vessel Q contains oil and water in two visible layers, Vessel R contains uniformly mixed air, and Vessel S contains muddy water that settles on standing. The pair that represents homogeneous solutions is:
ⓐ. P and Q
ⓑ. R and S
ⓒ. Q and S
ⓓ. P and R
Correct Answer: P and R
Explanation: Sugar dissolved completely in water forms a homogeneous liquid solution because the dissolved particles are uniformly distributed. Air is also treated as a homogeneous gaseous solution because its gaseous components are uniformly mixed on the ordinary scale. Oil and water form separate layers and therefore do not constitute a homogeneous solution. Muddy water contains suspended particles that may settle on standing, so it is not a true solution. Homogeneity and stable particle-level distribution, rather than the physical state of the mixture, identify P and R as solutions.
12. In a binary solution containing components \(P\) and \(Q\), the word binary tells us that the solution contains:
ⓐ. one solute and no solvent
ⓑ. exactly \(2\) components
ⓒ. two solvents and no solute
ⓓ. \(2\,mol\) of each component
Correct Answer: exactly \(2\) components
Explanation: A binary solution contains exactly two components. These two components are usually described as solute and solvent, although the exact identification depends on the situation. The word binary refers to the number of components, not to equal mole amounts or equal masses. A solution may be binary even if one component is present in a much smaller amount than the other. The number \(2\) in binary solution is a component count, not a concentration value.
13. In a solution prepared by dissolving a small amount of iodine in alcohol, alcohol is most suitably called the:
ⓐ. solute
ⓑ. solvent
ⓒ. precipitate
ⓓ. residue
Correct Answer: solvent
Explanation: The solvent is the component that provides the dissolving medium and is usually present in larger amount. In iodine solution in alcohol, iodine is the dissolved component, so it is the solute. Alcohol acts as the medium in which iodine is distributed uniformly. The solvent need not be water; many liquid solutions use organic liquids as solvents. Calling alcohol a precipitate or residue would be inappropriate because it is part of the homogeneous liquid solution.
14. The statement “the solvent is always the component present in larger amount” should be used carefully because:
ⓐ. every homogeneous mixture contains no identifiable solvent
ⓑ. molar mass alone determines the solvent in every mixture
ⓒ. phase and accepted convention may also identify the solvent
ⓓ. the smaller component must always be treated as the solvent
Correct Answer: phase and accepted convention may also identify the solvent
Explanation: In many simple solutions, the solvent is the component present in larger amount. This is a useful rule for ordinary cases such as salt dissolved in water. However, some cases are better understood by the phase of the final solution or by chemical convention. For example, in a gas dissolved in a liquid, the liquid is usually treated as the solvent because it determines the phase of the solution. The smaller-amount rule is helpful, but it should not be turned into an absolute law for every mixture.
15. A mixture contains \(CO_2\) dissolved in water under pressure. For the usual description of this solution, the solute and solvent are respectively:
ⓐ. \(CO_2\) and water
ⓑ. water and \(CO_2\)
ⓒ. \(CO_2\) and \(CO_2\)
ⓓ. water and water
Correct Answer: \(CO_2\) and water
Explanation: In carbonated water, \(CO_2\) is the gas dissolved in the liquid medium. Therefore \(CO_2\) is treated as the solute and water as the solvent. The final solution is liquid because the solvent phase is liquid. This example also shows that the solute is not always a solid; gases can also act as solutes. The phase of the solution is commonly guided by the solvent, not by the dissolved gas.
16. In the notation of a binary solution, components \(A\) and \(B\) are used. The safest interpretation of \(A\) and \(B\) at this stage is that they are:
ⓐ. two solvents present in equal amounts
ⓑ. the two components of the solution
ⓒ. two gaseous components of the solution
ⓓ. two substances having equal molar masses
Correct Answer: the two components of the solution
Explanation: In binary solution language, \(A\) and \(B\) are labels for the two components. They may later be used in formulas for mole fraction, vapour pressure, or other properties. The labels do not automatically tell which component has larger mass or which one is the solvent. They also do not require both components to be gases, liquids, or solids. Symbol labels should be read as component labels unless the question gives extra information.
17. A homogeneous mixture contains nitrogen, oxygen, argon, and carbon dioxide. Based on component count, it is best described as:
ⓐ. a binary gaseous solution
ⓑ. a four-component pure gas
ⓒ. a multicomponent gaseous solution
ⓓ. a single-component gaseous mixture
Correct Answer: a multicomponent gaseous solution
Explanation: The mixture contains more than two components, so it is not binary. Air is homogeneous on the ordinary scale and is treated as a gaseous solution. Because it contains nitrogen, oxygen, argon, carbon dioxide, and other minor gases, it is a multicomponent solution. A pure compound has a fixed chemical composition and is not formed by physically mixing several gases in variable proportions. The word multicomponent focuses on the number of substances present in the solution.
18. Match the example in Column I with the most suitable solute-solvent description in Column II.
| Column I | Column II |
|---|
| P. Salt in water | 1. Solid solute in liquid solvent |
| Q. \(CO_2\) in water | 2. Gas solute in liquid solvent |
| R. Ethanol in water | 3. Liquid solute in liquid solvent |
| S. Air | 4. Gas components in a gaseous solution |
ⓐ. P-4, Q-2, R-1, S-3
ⓑ. P-2, Q-1, R-3, S-4
ⓒ. P-1, Q-3, R-2, S-4
ⓓ. P-1, Q-2, R-3, S-4
Correct Answer: P-1, Q-2, R-3, S-4
Explanation: Salt in water is a common solid-in-liquid solution, so P matches 1. \(CO_2\) dissolved in water is gas-in-liquid, so Q matches 2. Ethanol mixed with water is a liquid-in-liquid solution, so R matches 3. Air is a gaseous solution because its main components are gases mixed uniformly. The classification uses the physical states of solute and solvent, not whether the example is familiar or aqueous.
19. The phase of a solution is generally decided by the:
ⓐ. phase of the solute before mixing
ⓑ. component present in the smaller amount
ⓒ. molar mass of the dissolved component
ⓓ. phase of the solvent
Correct Answer: phase of the solvent
Explanation: In ordinary classification of solutions, the phase of the solution is generally determined by the solvent. If a gas is dissolved in a liquid solvent, the solution is treated as a liquid solution. If solids are mixed uniformly in a solid solvent, the solution is treated as a solid solution. The solute may have a different physical state before dissolving. This is why soda water is described as a liquid solution even though one dissolved component is a gas.
20. Hydrogen absorbed in palladium is commonly classified as:
ⓐ. gas in solid solution
ⓑ. solid in gas solution
ⓒ. liquid in solid solution
ⓓ. gas in liquid solution
Correct Answer: gas in solid solution
Explanation: Hydrogen is the gaseous component and palladium is the solid medium. When hydrogen is absorbed in palladium, the solution is classified as gas in solid. The final phase is associated with the solid palladium, which acts as the solvent-like medium. This example is useful because it shows that solutions are not limited to liquids. A gas can be a solute even when the final solution is solid.
