201. Two gases \( X \) and \( Y \) obey Henry’s law in the same solvent at the same temperature. If \( K_H(X) > K_H(Y) \), then at the same partial pressure
ⓐ. gas \( X \) will have larger mole fraction in solution
ⓑ. both gases will have equal mole fraction in solution
ⓒ. gas \( Y \) will have larger mole fraction in solution
ⓓ. gas \( Y \) will have zero mole fraction in solution
Correct Answer: gas \( Y \) will have larger mole fraction in solution
Explanation: Since \( x = \frac{p}{K_H} \), the gas with smaller Henry’s law constant has the larger mole fraction in solution at the same pressure. Therefore, gas \( Y \) is more soluble than gas \( X \) here.
202. Why are soft drinks filled under high pressure?
ⓐ. Higher pressure increases the solubility of the gas in the liquid.
ⓑ. Higher pressure decreases the molarity of the drink.
ⓒ. Higher pressure converts the dissolved gas into a non-volatile solute.
ⓓ. Higher pressure lowers the temperature of the liquid automatically.
Correct Answer: Higher pressure increases the solubility of the gas in the liquid.
Explanation: Henry’s law states that gas solubility in a liquid increases with the pressure of that gas above the liquid. Filling under high pressure allows more gas to remain dissolved in the drink.
203. Which statement correctly distinguishes Henry’s law from Raoult’s law?
ⓐ. Henry’s law applies only to solid solutions, while Raoult’s law applies to gases.
ⓑ. Henry’s law uses mass percentage, while Raoult’s law uses molality.
ⓒ. Henry’s law is valid only at very high temperatures, while Raoult’s law is not.
ⓓ. Henry’s law concerns dissolved gas; Raoult’s law concerns vapour pressure.
Correct Answer: Henry’s law concerns dissolved gas; Raoult’s law concerns vapour pressure.
Explanation: Henry’s law relates the pressure of a gas to its solubility in a liquid. Raoult’s law, on the other hand, relates the partial vapour pressure of a liquid component to its mole fraction in a liquid solution.
204. At temperature \( T_1 \), a gas has \( K_H = 3.0 \times 10^4,\text{bar} \) in water. At temperature \( T_2 \), it has \( K_H = 6.0 \times 10^4,\text{bar} \). Which statement is correct?
ⓐ. The gas is more soluble at \( T_2 \).
ⓑ. The gas is less soluble at \( T_2 \).
ⓒ. The gas has equal solubility at both temperatures.
ⓓ. No comparison is possible from \( K_H \) values.
Correct Answer: The gas is less soluble at \( T_2 \).
Explanation: At a fixed pressure, Henry’s law gives \( x = \frac{p}{K_H} \). A larger value of \( K_H \) means a smaller mole fraction in solution, so the gas is less soluble at \( T_2 \).
205. In Henry’s law expression \( p = K_H x \), what does a larger value of \( K_H \) indicate for a gas in a given solvent at the same temperature?
ⓐ. Greater mole fraction of the gas at a given pressure
ⓑ. Lower solubility of the gas at a given pressure
ⓒ. No relation with the solubility of the gas
ⓓ. Complete dissociation of the gas in the solvent
Correct Answer: Lower solubility of the gas at a given pressure
Explanation: From Henry’s law, \( x = \frac{p}{K_H} \). For the same pressure, a larger value of \( K_H \) gives a smaller mole fraction of dissolved gas. Therefore, the gas is less soluble.
206. A gas obeys Henry’s law in water at a certain temperature. If \( K_H = 3.0 \times 10^4,\text{bar} \) and the partial pressure of the gas is \( 1.2,\text{bar} \), the mole fraction of the gas in water is
ⓐ. \( 4.0 \times 10^{-5} \)
ⓑ. \( 4.0 \times 10^{-4} \)
ⓒ. \( 2.5 \times 10^{-5} \)
ⓓ. \( 3.6 \times 10^{-5} \)
Correct Answer: \( 4.0 \times 10^{-5} \)
Explanation: \( \textbf{Given:} \)
\[
K_H = 3.0 \times 10^4,\text{bar}
\]
Partial pressure, \( p = 1.2,\text{bar} \)
\( \textbf{Required:} \)
Mole fraction of gas, \( x \)
\( \textbf{Relevant formula:} \)
\[
p = K_H x
\]
\( \textbf{Why this formula applies:} \)
Henry’s law relates the pressure of a gas above the liquid to its mole fraction in solution.
\( \textbf{Rearrange the formula:} \)
\[
x = \frac{p}{K_H}
\]
\( \textbf{Substitution:} \)
\[
x = \frac{1.2}{3.0 \times 10^4}
\]
\( \textbf{Intermediate simplification:} \)
\[
x = \frac{1.2}{3.0} \times 10^{-4}
\]
\[
x = 0.4 \times 10^{-4}
\]
\( \textbf{Final simplification:} \)
\[
x = 4.0 \times 10^{-5}
\]
\( \textbf{Unit:} \)
Mole fraction is unitless.
\( \textbf{Final Answer:} \)
Mole fraction of gas \( = 4.0 \times 10^{-5} \)
207. Why is high-pressure oxygen provided to a patient when a larger amount of oxygen is required to dissolve in blood?
ⓐ. Higher pressure increases oxygen solubility in blood.
ⓑ. Higher pressure always decreases the Henry’s law constant to zero.
ⓒ. Higher pressure converts oxygen into a non-volatile solute.
ⓓ. Higher pressure removes nitrogen completely from the blood.
Correct Answer: Higher pressure increases oxygen solubility in blood.
Explanation: Henry’s law states that the solubility of a gas increases with its pressure above the liquid. So increasing oxygen pressure helps more oxygen dissolve in blood.
208. For a gas in a liquid at a fixed temperature, the mole fraction of the dissolved gas is \( 3.0 \times 10^{-5} \) when the partial pressure is \( 0.90,\text{bar} \). The value of \( K_H \) is
ⓐ. \( 2.7 \times 10^{-5},\text{bar} \)
ⓑ. \( 3.0 \times 10^4,\text{bar} \)
ⓒ. \( 2.7 \times 10^4,\text{bar} \)
ⓓ. \( 9.0 \times 10^4,\text{bar} \)
Correct Answer: \( 3.0 \times 10^4,\text{bar} \)
Explanation: \( \textbf{Given:} \)
Partial pressure, \( p = 0.90,\text{bar} \)
Mole fraction, \( x = 3.0 \times 10^{-5} \)
\( \textbf{Required:} \)
Henry’s law constant, \( K_H \)
\( \textbf{Relevant formula:} \)
\[
p = K_H x
\]
\( \textbf{Why this formula applies:} \)
Henry’s law directly connects pressure and mole fraction.
\( \textbf{Rearrange the formula:} \)
\[
K_H = \frac{p}{x}
\]
\( \textbf{Substitution:} \)
\[
K_H = \frac{0.90}{3.0 \times 10^{-5}}
\]
\( \textbf{Intermediate simplification:} \)
\[
K_H = \frac{0.90}{3.0} \times 10^5
\]
\[
K_H = 0.30 \times 10^5
\]
\( \textbf{Final simplification:} \)
\[
K_H = 3.0 \times 10^4,\text{bar}
\]
\( \textbf{Unit:} \)
Since \( x \) is unitless and \( p \) is in bar, \( K_H \) is in bar.
\( \textbf{Final Answer:} \)
\[
K_H = 3.0 \times 10^4,\text{bar}
\]
209. Which statement correctly explains why a soft drink loses fizz more quickly when left open for a long time?
ⓐ. The gas becomes chemically bound to the liquid after opening.
ⓑ. Reduced pressure above the liquid lets dissolved gas escape.
ⓒ. The liquid becomes saturated with air and therefore stops obeying Henry’s law.
ⓓ. The bottle volume increases, so mole fraction of the gas becomes 1.
Correct Answer: Reduced pressure above the liquid lets dissolved gas escape.
Explanation: Once the bottle is opened, the pressure of the dissolved gas above the liquid decreases greatly. As a result, the dissolved gas becomes less soluble and gradually escapes.
210. A gas has mole fraction \( 2.0 \times 10^{-5} \) in a liquid at a pressure of \( 1.0,\text{bar} \). If the temperature remains constant and the pressure is raised to \( 2.5,\text{bar} \), the new mole fraction will be
ⓐ. \( 8.0 \times 10^{-6} \)
ⓑ. \( 2.0 \times 10^{-5} \)
ⓒ. \( 3.0 \times 10^{-5} \)
ⓓ. \( 5.0 \times 10^{-5} \)
Correct Answer: \( 5.0 \times 10^{-5} \)
Explanation: \( \textbf{Given:} \)
Initial mole fraction, \( x_1 = 2.0 \times 10^{-5} \)
Initial pressure, \( p_1 = 1.0,\text{bar} \)
Final pressure, \( p_2 = 2.5,\text{bar} \)
\( \textbf{Required:} \)
New mole fraction, \( x_2 \)
\( \textbf{Relevant principle:} \)
At constant temperature, Henry’s law gives \( p \propto x \)
\( \textbf{Why this principle applies:} \)
For the same gas-liquid system at fixed temperature, \( K_H \) remains constant.
So,
\[
\frac{p_1}{x_1} = \frac{p_2}{x_2}
\]
\( \textbf{Rearrange:} \)
\[
x_2 = x_1 \frac{p_2}{p_1}
\]
\( \textbf{Substitution:} \)
\[
x_2 = 2.0 \times 10^{-5} \times \frac{2.5}{1.0}
\]
\( \textbf{Intermediate simplification:} \)
\[
x_2 = 2.0 \times 2.5 \times 10^{-5}
\]
\( \textbf{Final simplification:} \)
\[
x_2 = 5.0 \times 10^{-5}
\]
\( \textbf{Unit:} \)
Mole fraction is unitless.
\( \textbf{Final Answer:} \)
\[
x_2 = 5.0 \times 10^{-5}
\]
211. Which quantity in Henry’s law represents the composition of the dissolved gas in the liquid phase?
ⓐ. \( x \)
ⓑ. \( p \)
ⓒ. \( K_H \)
ⓓ. \( \Delta p \)
Correct Answer: \( x \)
Explanation: In the relation \( p = K_H x \), the symbol \( x \) denotes the mole fraction of the gas dissolved in the liquid. It describes the composition of the gas in the solution phase.
212. For a certain gas in water, \( K_H \) changes from \( 2.0 \times 10^4,\text{bar} \) at \( T_1 \) to \( 5.0 \times 10^4,\text{bar} \) at \( T_2 \). At the same partial pressure, the ratio of mole fraction at \( T_2 \) to that at \( T_1 \) is
ⓐ. \( \frac{5}{2} \)
ⓑ. \( 1 \)
ⓒ. \( \frac{2}{5} \)
ⓓ. \( \frac{3}{5} \)
Correct Answer: \( \frac{2}{5} \)
Explanation: \( \textbf{Given:} \)
\[
K_H(T_1) = 2.0 \times 10^4,\text{bar}
\]
\[
K_H(T_2) = 5.0 \times 10^4,\text{bar}
\]
\( \textbf{Required:} \)
\( \frac{x_{T_2}}{x_{T_1}} \) at the same pressure
\( \textbf{Relevant formula:} \)
\[
x = \frac{p}{K_H}
\]
\( \textbf{Why this formula applies:} \)
At fixed pressure, mole fraction is inversely proportional to Henry’s law constant.
So,
\[
\frac{x_{T_2}}{x_{T_1}} = \frac{p/K_H(T_2)}{p/K_H(T_1)}
\]
\( \textbf{Intermediate simplification:} \)
\[
\frac{x_{T_2}}{x_{T_1}} = \frac{K_H(T_1)}{K_H(T_2)}
\]
\( \textbf{Substitution:} \)
\[
\frac{x_{T_2}}{x_{T_1}} = \frac{2.0 \times 10^4}{5.0 \times 10^4}
\]
\( \textbf{Final simplification:} \)
\[
\frac{x_{T_2}}{x_{T_1}} = \frac{2}{5}
\]
\( \textbf{Unit:} \)
This ratio is unitless.
\( \textbf{Final Answer:} \)
\[
\frac{x_{T_2}}{x_{T_1}} = \frac{2}{5}
\]
213. At constant temperature, the partial pressure of a gas above a solution is increased from \( 1.0,\text{bar} \) to \( 4.0,\text{bar} \). According to Henry’s law, the mole fraction of the dissolved gas will become
ⓐ. half of its initial value
ⓑ. twice of its initial value
ⓒ. four times its initial value
ⓓ. one-fourth of its initial value
Correct Answer: four times its initial value
Explanation: Henry’s law gives \( p = K_H x \) at constant temperature. Since \( K_H \) remains constant, the mole fraction \( x \) is directly proportional to pressure. Increasing the pressure from \( 1.0,\text{bar} \) to \( 4.0,\text{bar} \) makes the mole fraction four times.
214. Which statement correctly follows from Henry’s law for gases dissolved in liquids?
ⓐ. At a fixed temperature, lower pressure gives lower gas solubility.
ⓑ. At a fixed temperature, the dissolved gas always has mole fraction 1.
ⓒ. At a fixed temperature, gas solubility is independent of pressure.
ⓓ. At a fixed temperature, the pressure above the liquid decreases when more gas dissolves.
Correct Answer: At a fixed temperature, lower pressure gives lower gas solubility.
Explanation: Henry’s law states that the solubility of a gas is directly proportional to the pressure of that gas above the liquid at constant temperature. Therefore, lowering the pressure lowers the dissolved amount.
215. Vapour pressure of a pure liquid is best defined as the
ⓐ. pressure exerted by air dissolved in the liquid
ⓑ. external pressure required to boil the liquid
ⓒ. pressure exerted by all gases above an open liquid
ⓓ. pressure of vapour in equilibrium with its liquid
Correct Answer: pressure of vapour in equilibrium with its liquid
Explanation: Vapour pressure is an equilibrium property. It is the pressure exerted by the vapour when the rate of evaporation equals the rate of condensation in a closed container.
216. In a closed vessel containing a pure liquid, vapour pressure becomes constant when
ⓐ. all the liquid has evaporated
ⓑ. rates become equal
ⓒ. condensation stops completely
ⓓ. the liquid starts boiling
Correct Answer: rates become equal
Explanation: A constant vapour pressure is established when dynamic equilibrium is reached. At that stage, molecules continue to evaporate and condense, but the two rates are equal.
217. Which statement about a volatile liquid is correct?
ⓐ. It has appreciable vapour pressure.
ⓑ. It has zero vapour pressure below its boiling point.
ⓒ. It cannot exist in the liquid state in a closed vessel.
ⓓ. It always behaves as a non-ideal solution.
Correct Answer: It has appreciable vapour pressure.
Explanation: A volatile liquid evaporates readily and therefore shows appreciable vapour pressure even at ordinary temperatures. This escaping tendency is the key feature of volatility.
218. Why does vapour pressure of a pure liquid generally increase with rise in temperature?
ⓐ. The liquid becomes chemically less stable.
ⓑ. The molar mass of the vapour decreases.
ⓒ. External pressure always falls on heating.
ⓓ. More molecules escape into the vapour phase.
Correct Answer: More molecules escape into the vapour phase.
Explanation: When temperature increases, the kinetic energy of liquid molecules increases. As a result, a larger number of molecules can overcome intermolecular attractions and enter the vapour phase, raising the vapour pressure.
219. Which statement is incorrect for a liquid-vapour equilibrium in a closed vessel?
ⓐ. Molecules keep moving from liquid to vapour and from vapour to liquid.
ⓑ. The vapour pressure at equilibrium depends on the nature of the liquid.
ⓒ. The amount of vapour becomes zero at equilibrium.
ⓓ. The equilibrium is dynamic in nature.
Correct Answer: The amount of vapour becomes zero at equilibrium.
Explanation: At equilibrium, vapour is present above the liquid and exerts a definite vapour pressure. The system is dynamic because evaporation and condensation continue simultaneously.
220. Which pair will generally have the higher vapour pressure at the same temperature?
ⓐ. The liquid with stronger intermolecular attraction
ⓑ. The liquid with lower escaping tendency
ⓒ. The liquid with slower evaporation rate
ⓓ. The liquid with weaker intermolecular attraction
Correct Answer: The liquid with weaker intermolecular attraction
Explanation: When intermolecular attractions are weaker, molecules escape more easily from the liquid surface into the vapour phase. Therefore, the liquid shows higher vapour pressure at the same temperature.
