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In this Class 12 Chemistry topic from Chapter 01: Solutions, students learn how vapour pressure arises from the dynamic equilibrium between evaporation and condensation in a liquid. The topic explains the effect of temperature and the presence of a non-volatile solute, including lowering of vapour pressure. Students also connect vapour pressure with mole fraction through Raoult’s law and examine how ideal and non-ideal solutions differ, using equations and basic numerical applications.
TOPIC PRACTICE
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Up to 25 questions from this page. Select your focus, then start.
25 questions
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Easy · Level 2View options
Lower
Higher
It must be the same
Zero
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It decreases
It increases
It always remains one
It becomes infinite
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Half
Double
Four times
Zero
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It is almost none
It is very high
It is always more than pure solvent
It depends only on colour
Easy · Level 2View options
Total pressure is the sum of partial pressures
Total pressure is always zero
Total pressure depends only on colour
Total pressure always decreases
Easy · Level 2View options
It increases
It decreases
It becomes zero
It depends on colour
Easy · Level 2View options
Positive deviation
Negative deviation
No deviation
Osmotic deviation
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Negative deviation
Positive deviation
Perfect ideal behaviour
Sublimation
Easy · Level 2View options
No, it mainly depends on temperature and nature
Yes, double amount gives double pressure
Yes, less amount gives zero pressure
It depends only on container colour
Easy · Level 2View options
A slightly larger or smaller amount of liquid
Nature of the liquid
Temperature
Intermolecular attraction
Easy · Level 2View options
Vapour escapes out
Liquid has no particles
Temperature is always zero
Liquid never forms vapour
Easy · Level 2View options
It decreases
It increases
It becomes infinite
It depends on colour
Easy · Level 2View options
A non-volatile solute is dissolved
Name of solvent has changed
Colour of container has changed
Liquid has no temperature
Easy · Level 2View options
More particles become vapour as temperature increases
Particle motion always stops on heating
Vapour pressure is always infinite on cooling
Temperature has no relation with vapour pressure
Easy · Level 2View options
A non-volatile solute lowers vapour pressure
A non-volatile solute increases vapour pressure
Vapour pressure has no relation with temperature
Raoult's law is based on colour
Easy · Level 2View options
Pressure exerted by vapour at equilibrium
Pressure exerted by liquid on the vessel
Weight of the solution
Volume of the liquid
Easy · Level 2View options
It decreases
It increases
It becomes double
It always becomes zero
Easy · Level 2View options
Lower
Higher
Equal
Infinite
Easy · Level 2View options
Volatile components
Only non-volatile solute
Only container
Only colour of liquid
Easy · Level 2View options
Solvent molecules
Solute molecules
Both in equal amount
No molecule
Easy · Level 2View options
Vapour pressure of the pure solvent
Mass of the solute
Colour of the solution
Only the pressure of the container
Easy · Level 2View options
Mole fraction
Mass
Density
Time
Easy · Level 2View options
Pure water
Sugar solution
Both equal
Neither
Easy · Level 2View options
Raoult's law
Hess's law
Newton's law of motion
Only Le Chatelier principle
Easy · Level 2View options
Add a non-volatile solute
Increase temperature
Keep solvent pure
Change container colour
Question 1EasyLevel 2
If one of two liquids has higher vapour pressure, how will its boiling point generally be?
Correct answer: A
Boiling begins when a liquid's vapour pressure equals the external pressure. At the same temperature, the liquid with higher vapour pressure is closer to this condition and reaches it after less heating. Therefore its boiling point is generally lower. The comparison assumes the same external pressure; boiling point is not necessarily zero or equal for both liquids.
What happens to the mole fraction of the solvent when a non-volatile solute is added to a solution?
Correct answer: A
The solvent mole fraction is n_solvent/(n_solvent + n_solute). Adding a nonzero amount of solute increases the denominator while the moles of solvent remain unchanged, so the fraction becomes less than its previous value. For pure solvent it is one, but after adding solute it is no longer one and can never become infinite.
If the mole fraction of the solvent is 0.5, according to Raoult's law, what fraction of the pure solvent vapour pressure will the solution vapour pressure be?
Correct answer: A
Raoult’s law states P_solution = X_solvent P°_solvent. Therefore, when X_solvent = 0.5, P_solution = 0.5P°_solvent, which means the solution vapour pressure is one-half of the pure solvent vapour pressure. It is not double or four times; nor is it zero, because the solvent still has a non-zero mole fraction and can still evaporate.
What is the contribution of a non-volatile solute to vapour pressure?
Correct answer: A
A non-volatile solute has an extremely low tendency to enter the vapour phase under the conditions considered. Hence its own partial vapour pressure is taken as negligible, and the measured vapour pressure is essentially due to the solvent. Its presence still lowers solvent vapour pressure by lowering the solvent mole fraction.
What is the simple meaning of Dalton's law for total vapour pressure?
Correct answer: A
Dalton's law states that, for a mixture of non-reacting gases behaving approximately ideally, the total pressure equals the arithmetic sum of the partial pressures exerted by all components. In a vapour mixture, each volatile liquid contributes one partial pressure. The law does not say that total pressure is always zero or always decreasing.
If the mole fraction of a component increases in an ideal solution, what happens to its partial vapour pressure?
Correct answer: A
For each volatile component in an ideal solution, Raoult's law gives P_i = X_i P°_i. At a fixed temperature, the pure-component pressure P°_i is constant, so increasing X_i increases that component's partial vapour pressure in direct proportion. The pressure does not become zero, and colour has no role in this quantitative relation.
If actual vapour pressure is greater than the value expected from Raoult's law, which deviation is shown?
Correct answer: A
Raoult's law gives the ideal reference vapour pressure. If experimental or actual pressure lies above that reference, the difference is called positive deviation. It indicates that molecules escape more readily than in the ideal model, often because unlike intermolecular attractions are weaker. A lower actual value would indicate negative deviation.
If the actual vapour pressure is less than the value expected from Raoult's law, which deviation is shown?
Correct answer: A
Deviation is judged by comparing the observed vapour pressure with the Raoult-law value. If the observed pressure is lower, the mixture shows negative deviation. This usually indicates stronger unlike-molecule attractions, which hold molecules in the liquid more effectively. Positive deviation would mean a pressure higher than the ideal prediction.
Does vapour pressure mainly depend on the amount of liquid if liquid is present and temperature is constant?
Correct answer: A
For a pure liquid at fixed temperature, if some liquid remains and liquid-vapour equilibrium can be established, the equilibrium vapour pressure is an intensive property. Changing the amount changes the time or amount of vapour needed for equilibrium, not the final pressure. If all liquid evaporates, this conclusion no longer applies.
At constant temperature, equilibrium vapour pressure of a pure liquid does not depend on which factor?
Correct answer: A
When some liquid remains present in a closed system, equilibrium vapour pressure is fixed by temperature and the nature of the liquid. Changing the amount slightly changes how long equilibrium takes to establish, but not the final pressure. If all liquid evaporates, the condition changes; the option is understood to mean a change while liquid remains present.
Why is it difficult to measure vapour pressure of a liquid kept in an open container?
Correct answer: A
A measurable equilibrium vapour pressure requires vapour to accumulate above the liquid until evaporation and condensation rates become equal. In an open container, vapour continuously diffuses away, so the required closed-system equilibrium is not established. The liquid still contains particles and can evaporate; its temperature is not forced to zero.
When vapour pressure decreases, what happens to the tendency of a liquid to form vapour?
Correct answer: A
At a specified temperature, vapour pressure reflects the escaping tendency of molecules from a liquid into the gas phase. A lower equilibrium vapour pressure means that fewer molecules can escape sufficiently to maintain the vapour phase. Therefore the liquid is less volatile and has a lower tendency to form vapour. Colour does not determine this property.
If vapour pressure of a solution is lower than that of pure solvent, what is the most likely reason?
Correct answer: A
At the same temperature, adding a non-volatile solute lowers the solvent mole fraction. Raoult's law then gives a lower solvent vapour pressure than for the pure solvent. Changing a name or container colour has no thermodynamic effect, and every liquid has a temperature; the decrease is not explained by those options.
Which statement correctly explains the relation between vapour pressure and temperature?
Correct answer: A
Increasing temperature raises the kinetic-energy distribution of liquid molecules. A greater fraction then has enough energy to escape the surface, so the equilibrium vapour pressure rises. Cooling has the opposite general effect. Heating does not stop molecular motion, and vapour pressure is neither independent of temperature nor infinite on cooling.
What is the most basic exam idea in the vapour pressure topic?
Correct answer: A
The central idea is that a non-volatile solute lowers the solvent mole fraction and therefore lowers the solvent vapour pressure: p = x_solvent p°. This result leads to elevation of boiling point and depression of freezing point. Temperature still affects vapour pressure, and Raoult's law is a mole-fraction relation, not a colour relation.
When a liquid is placed in a closed vessel, molecules evaporate while vapour molecules condense back. At dynamic equilibrium, the rate of evaporation equals the rate of condensation. The pressure exerted by the vapour at that equilibrium, at a specified temperature, is called vapour pressure. It is not the liquid's weight, volume or wall pressure.
What happens to the vapour pressure of a solvent when a non-volatile solute is added?
Correct answer: A
A non-volatile solute does not contribute appreciably to the vapour phase. Its presence lowers the mole fraction and escaping tendency of the solvent. For an ideal solution, Raoult’s law gives psolvent = xsolvent p°solvent; since xsolvent is less than one, the solvent vapour pressure is lower than that of the pure solvent.
Compared with the pure solvent, how is the vapour pressure of a solution containing a non-volatile solute?
Correct answer: A
A non-volatile solute does not contribute appreciably to the vapour above the solution. Its presence reduces the mole fraction and the escaping tendency of solvent molecules at the liquid surface. For an ideal solution, Raoult’s law gives p = x_solvent p°_solvent, and because x_solvent is less than 1, p is lower than the vapour pressure of the pure solvent. Therefore A is correct.
Who contributes to the total vapour pressure of a solution?
Correct answer: A
A component contributes to vapour pressure only if its molecules can enter the vapour phase in appreciable amount. Each volatile component exerts a partial vapour pressure, and the total pressure is the sum of these partial pressures. A genuinely non-volatile solute contributes no significant direct vapour pressure; the container and colour do not contribute.
If only the solvent is volatile and the solute is non-volatile, what will mainly be present in the vapour?
Correct answer: A
The volatile solvent can escape from the liquid and establish a vapour phase. A non-volatile solute has negligible tendency to enter that phase under the stated conditions. Therefore, the vapour consists mainly of solvent molecules. The two components cannot be assumed equal, and vapour is not completely absent while solvent remains present.
In Raoult’s law for a solution containing a volatile solvent, the relation is p_solvent = x_solvent p°_solvent. Here p_solvent is the partial vapour pressure of the solvent in the solution, while p°_solvent is the vapour pressure of the pure solvent at the same temperature. Thus p° is a pressure reference, not the solute mass, solution colour, or pressure of the container. Option A is correct.
In Raoult’s law, x represents the mole fraction of the component whose partial vapour pressure is being calculated. For a solvent, p = xsolvent p°solvent. Mole fraction is the component’s moles divided by the total moles of all components, so it is dimensionless; mass, density and time have different meanings and units.
At the same temperature, which has higher vapour pressure: pure water or sugar solution in water?
Correct answer: A
At the same temperature, dissolved sugar is treated as a non-volatile solute. It lowers the mole fraction and escaping tendency of water, so the water vapour pressure in the sugar solution is below the vapour pressure of pure water. Therefore pure water has the higher value; the result assumes the solution is compared at the same temperature.
In terms of vapour pressure, an ideal solution obeys which law?
Correct answer: A
For an ideal solution, each component obeys Raoult's law: its partial vapour pressure equals its mole fraction multiplied by the vapour pressure of the pure component. Ideal solutions have similar intermolecular interactions and show no significant heat or volume change on mixing. Hess's law and Newton's laws address different subjects.
Which method can lower the vapour pressure of a solvent?
Correct answer: A
Adding a non-volatile solute reduces the solvent mole fraction and therefore lowers its partial vapour pressure according to Raoult's law. Increasing temperature generally raises vapour pressure, while keeping the solvent pure does not produce lowering. Changing container colour has no thermodynamic effect on the equilibrium vapour pressure.
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