101. A solubility value is reported as \(20\,g\) solute per \(100\,g\) water at \(300\,K\). If \(10\,g\) solute is mixed with \(100\,g\) water at \(300\,K\), the solution formed is:
ⓐ. saturated with \(10\,g\) undissolved solute
ⓑ. saturated with \(20\,g\) undissolved solute
ⓒ. unsaturated if all \(10\,g\) dissolves
ⓓ. pure solvent because less solute was added
Correct Answer: unsaturated if all \(10\,g\) dissolves
Explanation: \( \textbf{Solubility limit:} \) At \(300\,K\), up to \(20\,g\) solute can dissolve in \(100\,g\) water.
\( \textbf{Solute added:} \) Only \(10\,g\) solute is added to \(100\,g\) water.
\( \textbf{Comparison:} \)
\[
10\,g \lt 20\,g
\]
\( \textbf{Interpretation:} \) The solution is below the maximum dissolving limit if all the added solute dissolves.
\( \textbf{Classification:} \) It is unsaturated because it can still dissolve more solute at \(300\,K\).
\( \textbf{Final answer:} \) The solution is unsaturated if all \(10\,g\) dissolves. Adding less than the solubility limit does not make the liquid pure solvent.
102. A small passage describes a beaker: At \(298\,K\), a solution contains dissolved sugar and a few sugar crystals at the bottom. After some time, the size of the crystals remains constant, although particles continue to leave and join the crystal surface. This situation represents:
ⓐ. dynamic equilibrium in a saturated solution
ⓑ. complete absence of dissolution
ⓒ. formation of a new compound from sugar and water
ⓓ. a gaseous solution under pressure
Correct Answer: dynamic equilibrium in a saturated solution
Explanation: The presence of crystals with a solution that no longer changes in concentration is typical of saturation. The passage also states that particles continue to leave and join the crystal surface, which indicates opposing processes. If the crystal size remains constant, the rates of dissolution and crystallisation are equal. This is dynamic equilibrium, not a static condition. Sugar and water remain physically mixed in solution rather than forming a new compound.
103. A supersaturated solution is best described as a solution that:
ⓐ. contains less solute than an unsaturated solution at the same temperature
ⓑ. contains more dissolved solute than the equilibrium solubility permits
ⓒ. contains solute but no molecules of the solvent
ⓓ. forms only when pressure on a solid solute is decreased
Correct Answer: contains more dissolved solute than the equilibrium solubility permits
Explanation: A supersaturated solution contains more dissolved solute than would normally remain in a saturated solution at the same temperature. Such a solution is not a stable equilibrium state. A small crystal of solute, shaking, or dust can start crystallisation. The excess dissolved solute then separates until the solution returns closer to saturation. This idea is different from an unsaturated solution, which can still dissolve more solute under the same conditions.
104. A hot saturated solution of \(KNO_3\) is cooled carefully without disturbance and remains clear, even though its concentration is higher than the normal solubility at the lower temperature. The solution is:
ⓐ. dilute only
ⓑ. supersaturated
ⓒ. ideal binary volatile
ⓓ. isotonic
Correct Answer: supersaturated
Explanation: Cooling a hot saturated solution can sometimes produce a clear solution that contains more solute than the lower-temperature solubility allows. This is a supersaturated solution. It is clear because crystallisation has not yet started. It is unstable and can crystallize when disturbed or seeded with a crystal. The term does not describe osmotic pressure or ideal vapour-pressure behaviour.
105. For many solids dissolving in liquids, an endothermic dissolution process is generally favoured by:
ⓐ. lowering temperature, which removes heat from the system
ⓑ. raising pressure, because solid solutes are highly compressible
ⓒ. removing solvent, which shifts the thermal equilibrium
ⓓ. raising temperature, which favours the heat-absorbing direction
Correct Answer: raising temperature, which favours the heat-absorbing direction
Explanation: In an endothermic dissolution process, heat is absorbed as the solute dissolves. Increasing temperature supplies heat to the system, so the equilibrium generally shifts toward greater dissolution. This often increases the solubility of such solids in liquids. Pressure usually has little effect on the solubility of solids in liquids because solids and liquids are not highly compressible. The temperature effect should be linked to the heat change of dissolution, not treated as the same for every solid.
106. A salt has solubility \(30\,g\) per \(100\,g\) water at \(300\,K\) and \(45\,g\) per \(100\,g\) water at \(320\,K\). A saturated solution prepared with \(100\,g\) water at \(300\,K\) is heated to \(320\,K\). The extra mass of salt that can dissolve is:
ⓐ. \(15\,g\)
ⓑ. \(30\,g\)
ⓒ. \(45\,g\)
ⓓ. \(75\,g\)
Correct Answer: \(15\,g\)
Explanation: \( \textbf{Solubility at }300\,K\textbf{:} \) \(30\,g\) salt dissolves in \(100\,g\) water.
\( \textbf{Solubility at }320\,K\textbf{:} \) \(45\,g\) salt dissolves in \(100\,g\) water.
\( \textbf{Initial dissolved mass at saturation:} \)
\[
30\,g
\]
\( \textbf{Maximum dissolved mass after heating:} \)
\[
45\,g
\]
\( \textbf{Extra mass that can dissolve:} \)
\[
45\,g-30\,g=15\,g
\]
\( \textbf{Final answer:} \) The extra amount that can dissolve is \(15\,g\). The comparison must be made for the same \(100\,g\) water basis at the two temperatures.
107. Dissolution of a certain solid in water is exothermic. On increasing temperature, its solubility generally:
ⓐ. increases because higher temperature favours every dissolution
ⓑ. remains fixed because solvent nature controls it completely
ⓒ. increases only when pressure is raised at the same time
ⓓ. decreases because added heat opposes the exothermic direction
Correct Answer: decreases because added heat opposes the exothermic direction
Explanation: In an exothermic dissolution process, heat is released when the solid dissolves. Raising temperature is like adding heat to the system, so the equilibrium tends to oppose this change by reducing dissolution. Therefore solubility generally decreases for such a solid as temperature increases. This is not a universal rule for all solids; the direction depends on the heat change of dissolution. The phrase “hot water dissolves all solids better” is too broad for solubility reasoning.
108. A data table for solid \(X\) in water is shown below.
| Temperature | Solubility of \(X\) |
|---|
| \(290\,K\) | \(52\,g\) per \(100\,g\) water |
| \(310\,K\) | \(46\,g\) per \(100\,g\) water |
| \(330\,K\) | \(40\,g\) per \(100\,g\) water |
The dissolution of \(X\) is most likely:
ⓐ. pressure-controlled, as expected for a dissolved gas
ⓑ. temperature-independent over the measured range
ⓒ. exothermic in the overall dissolution process
ⓓ. unable to dissolve in water at these temperatures
Correct Answer: exothermic in the overall dissolution process
Explanation: The solubility of \(X\) decreases as temperature increases. This pattern is consistent with an exothermic dissolution process, where heating generally reduces the amount that dissolves at equilibrium. The table does not show pressure dependence, and pressure is usually not a major factor for solids in liquids. The solid clearly dissolves because the table reports finite solubility values. The temperature trend is the clue for the likely heat effect of dissolution.
109. A student says, “Increasing temperature always increases the solubility of a solid in a liquid.” The best correction is:
ⓐ. endothermic dissolution always becomes less soluble on heating
ⓑ. the trend depends on the enthalpy change of dissolution
ⓒ. pressure controls the temperature trend for every solid
ⓓ. exothermic dissolution always becomes more soluble on heating
Correct Answer: the trend depends on the enthalpy change of dissolution
Explanation: Many solids show increased solubility with rise in temperature, but this is not a universal statement. If dissolution is endothermic, heating generally favours greater dissolution. If dissolution is exothermic, heating may decrease solubility. The nature of the solute-solvent system also matters. A reliable explanation must connect the temperature effect with the dissolution equilibrium instead of applying one fixed rule to every solid.
110. Assertion: Pressure change usually has a negligible effect on the solubility of solids in liquids.
Reason: Solids and liquids are much less compressible than gases.
ⓐ. 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 for solids dissolving in liquids under ordinary pressure changes. The Reason is also true because solids and liquids do not undergo large volume changes when pressure changes moderately. Since pressure effects are closely related to volume changes, the solubility of solids in liquids is usually not strongly pressure-controlled. This contrasts with gases in liquids, where pressure has a clear effect. The Reason gives the physical basis for the negligible pressure effect in solid-liquid systems.
111. A saturated solution at \(310\,K\) is cooled to \(290\,K\). The solubility of the solid is lower at \(290\,K\). The most likely observation is:
ⓐ. more solid dissolves into the solution
ⓑ. the solution becomes a gas-liquid solution
ⓒ. the amount of solvent increases by itself
ⓓ. some dissolved solid crystallises out
Correct Answer: some dissolved solid crystallises out
Explanation: At \(310\,K\), the solution contains the maximum dissolved solute for that temperature. If cooling lowers the solubility, the same amount of solvent can no longer hold all the solute that was dissolved before. The excess solute separates out as crystals. This does not mean the solvent amount has increased or that the solution has changed into a gas-liquid system. Crystallisation on cooling is expected when the lower-temperature solubility limit is smaller.
112. A gas is dissolved in a liquid at constant temperature. Increasing the pressure of the gas above the liquid generally:
ⓐ. decreases the solubility of the gas
ⓑ. converts the liquid solvent into a solid
ⓒ. has the same negligible effect as for solids in liquids
ⓓ. increases the solubility of the gas
Correct Answer: increases the solubility of the gas
Explanation: The solubility of a gas in a liquid generally increases when the pressure of that gas above the liquid is increased. Higher pressure means more gas particles strike the liquid surface and more enter the liquid phase. This behaviour is much more significant for gases than for solids because gases are highly affected by pressure. The statement assumes constant temperature and no chemical reaction changing the gas. This is the basic pressure trend behind sealed carbonated drinks.
113. A sealed carbonated drink contains dissolved \(CO_2\) under high pressure. When the bottle is opened, bubbles appear mainly because:
ⓐ. lower pressure allows dissolved \(CO_2\) to escape
ⓑ. opening cools the liquid instantly to \(0\,K\)
ⓒ. water decomposes chemically and produces \(CO_2\)
ⓓ. lower pressure increases the solubility of dissolved \(CO_2\)
Correct Answer: lower pressure allows dissolved \(CO_2\) to escape
Explanation: In a sealed carbonated drink, \(CO_2\) is kept dissolved by maintaining high pressure above the liquid. When the bottle is opened, the pressure above the liquid drops. The gas becomes less soluble at the lower pressure, so some dissolved \(CO_2\) escapes as bubbles. The water is not decomposing to produce \(CO_2\). The observation is a direct everyday example of gas solubility decreasing when gas pressure is reduced.
114. Warm water usually holds less dissolved gas than cold water because:
ⓐ. higher temperature strengthens gas-solvent attractions and favours dissolution
ⓑ. higher temperature helps gas molecules escape from the liquid
ⓒ. gas molecules move more slowly at higher temperature
ⓓ. heating increases the pressure that traps gas in the liquid
Correct Answer: higher temperature helps gas molecules escape from the liquid
Explanation: Gas solubility in liquids generally decreases as temperature increases. At higher temperature, dissolved gas molecules have greater tendency to escape from the liquid phase. This is why warm water tends to lose dissolved gases more readily than cold water. The trend is opposite to many endothermic solid-dissolution cases, so the type of solute matters. Temperature effects for gases should not be copied from a simple “hot water dissolves more” idea.
115. A cold bottle of soda is opened gently, and another similar warm bottle is opened gently at the same external pressure. More rapid gas escape is expected from the warm bottle mainly because:
ⓐ. \(CO_2\) changes into a solid at the warmer temperature
ⓑ. water ceases to act as a solvent when its temperature rises
ⓒ. \(CO_2\) is generally less soluble in warmer water
ⓓ. gas solubility becomes independent of pressure in warm water
Correct Answer: \(CO_2\) is generally less soluble in warmer water
Explanation: Dissolved gases are generally less soluble in liquids at higher temperature. In the warm soda bottle, \(CO_2\) has a greater tendency to escape from the liquid once the bottle is opened. Both bottles experience a pressure drop on opening, but the warm liquid holds less \(CO_2\) under the same external pressure. Water still acts as the solvent in soda water. The stronger bubbling is connected to lower gas solubility at the higher temperature.
116. A lake receives warm discharge water from a nearby plant. The dissolved oxygen level in the warmer region may decrease because:
ⓐ. oxygen is generally less soluble in water at higher temperature
ⓑ. oxygen becomes more soluble as water temperature rises
ⓒ. warming necessarily raises the oxygen partial pressure above water
ⓓ. gas-solvent attractions become stronger in warmer water
Correct Answer: oxygen is generally less soluble in water at higher temperature
Explanation: Dissolved oxygen is a gas dissolved in water. For gases in liquids, solubility usually decreases when temperature increases. Warm discharge water can therefore hold less dissolved oxygen than cooler water under similar pressure conditions. This can affect aquatic organisms that depend on dissolved oxygen. The trend comes from temperature-dependent gas solubility, not from a necessary rise in oxygen pressure or stronger gas-solvent attraction.
117. At \(300\,K\) and \(1\,bar\), a gas is moderately soluble in a liquid. The condition expected to give the highest gas solubility is:
ⓐ. \(320\,K\) and \(1\,bar\)
ⓑ. \(300\,K\) and \(0.5\,bar\)
ⓒ. \(320\,K\) and \(0.5\,bar\)
ⓓ. \(280\,K\) and \(2\,bar\)
Correct Answer: \(280\,K\) and \(2\,bar\)
Explanation: Gas solubility generally increases with pressure and decreases with temperature. To maximize solubility, the pressure should be higher and the temperature should be lower. Among the options, \(280\,K\) and \(2\,bar\) combines lower temperature with higher pressure. The conditions with \(320\,K\) reduce gas solubility because the temperature is higher. The condition \(300\,K\) and \(0.5\,bar\) also lowers solubility because the pressure is lower.
118. A comparison of solubility trends is shown below.
| System | Change applied | Usual effect |
|---|
| P. Gas in liquid | Increase pressure at constant temperature | Solubility increases |
| Q. Gas in liquid | Increase temperature at fixed pressure | Solubility usually decreases |
| R. Solid in liquid | Increase pressure moderately | Large solubility increase is usually expected |
| S. Endothermic solid dissolution | Increase temperature | Solubility generally increases |
The row that needs correction is:
ⓐ. P
ⓑ. Q
ⓒ. R
ⓓ. S
Correct Answer: R
Explanation: Row P is correct because gas solubility generally increases with pressure. Row Q is correct because gas solubility in liquids usually decreases when temperature rises. Row S is also correct because endothermic dissolution is generally favoured by heating. Row R is not consistent with the usual behaviour of solids in liquids, because moderate pressure changes generally have little effect on their solubility. The pressure-sensitive system in this comparison is gas in liquid, not solid in liquid.
119. Henry's law for a gas dissolved in a liquid at constant temperature is most suitably written as:
ⓐ. \(p=K_H+x\)
ⓑ. \(p=\frac{x}{K_H}\)
ⓒ. \(p=xK_H\)
ⓓ. \(p=\frac{K_H}{x}\)
Correct Answer: \(p=xK_H\)
Explanation: Henry's law states that the partial pressure of a gas above a solution is proportional to its mole fraction in the solution at constant temperature. The relation is commonly written as \(p=K_Hx\), which is the same as \(p=xK_H\). Here \(p\) is the partial pressure of the gas, \(x\) is its mole fraction in the solution, and \(K_H\) is Henry's law constant. The relation is not a sum or a reciprocal relation. The constant-temperature condition is essential because \(K_H\) changes with temperature.
120. Henry's law applies to a gas at a fixed temperature with \(K_H=8.0\times10^4\,kPa\). If the mole fraction of the gas in solution is \(2.0\times10^{-5}\), its partial pressure is:
ⓐ. \(0.16\,kPa\)
ⓑ. \(1.6\,kPa\)
ⓒ. \(16\,kPa\)
ⓓ. \(4.0\times10^9\,kPa\)
Correct Answer: \(1.6\,kPa\)
Explanation: \( \textbf{Known data:} \) \(K_H=8.0\times10^4\,kPa\), \(x=2.0\times10^{-5}\).
\( \textbf{Required quantity:} \) Partial pressure \(p\).
\( \textbf{Henry's law relation:} \)
\[
p=K_Hx
\]
\( \textbf{Reason for using it:} \) The gas obeys Henry's law at fixed temperature, and \(x\) is given.
\( \textbf{Substitution:} \)
\[
p=(8.0\times10^4)(2.0\times10^{-5})
\]
\( \textbf{Power simplification:} \)
\[
10^4\times10^{-5}=10^{-1}
\]
\( \textbf{Numerical calculation:} \)
\[
p=16.0\times10^{-1}=1.6\,kPa
\]
\( \textbf{Final answer:} \) The partial pressure is \(1.6\,kPa\). Since mole fraction is dimensionless, the pressure unit comes from \(K_H\).
