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In this Class 12 Chemistry topic from Chapter 01: Solutions, students learn how much of a solute can dissolve in a given amount of solvent under specific conditions. The topic explains saturated, unsaturated and supersaturated solutions, along with the factors that affect solubility, such as the nature of solute and solvent, temperature and pressure. Students also explore why gases behave differently from solids in solutions and apply these ideas to interpret solubility data and related chemical situations.
TOPIC PRACTICE
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Medium · Level 4View options
The one whose solubility changes greatly with temperature
The one whose solubility never changes
The one insoluble at every temperature
The one that chemically destroys the solvent
Medium · Level 4View options
Its Henry’s law constant will be lower
Its Henry’s law constant will be higher
Both constants must be zero
The constant will have no relation to solubility
Medium · Level 4View options
Some solute may separate out
The solution immediately becomes unsaturated
All solute becomes gas
Solubility becomes infinite
Medium · Level 4View options
Fractional crystallisation
Boiling into gas
Filtration only
Making pressure zero
Medium · Level 4View options
Pressure decreases slowly and dissolved gas comes out in a controlled way
Blood colour changes immediately
Density of water becomes zero
Gases dissolve more
Medium · Level 4View options
When attraction between ions and water molecules is strong
When ions remain completely away from water
When water has no polarity
When temperature is always zero
Medium · Level 4View options
Solubility decreases
Solubility increases
Solubility becomes infinite
There is no effect on solubility
Medium · Level 4View options
Solubility of the salt will decrease
Solubility of the salt will increase
Salt will completely become gas
Solvent will disappear
Medium · Level 4View options
It is related to the maximum equilibrium amount of ions
It tells only the colour of solvent
It measures gas pressure
It tells the shape of container
Medium · Level 4View options
A precipitate will form
The whole solution will become a gas
Solubility will become infinite
No ions will remain present
Medium · Level 4View options
The solution will be unsaturated
Precipitate must form
There will be no ions in solution
Salt will evaporate immediately
Medium · Level 4View options
A saturated solution
An unsaturated solution
A supersaturated solution in which precipitation cannot occur
A completely gaseous solution
Medium · Level 4View options
More soluble
Less soluble
Always insoluble
Always volatile
Medium · Level 4View options
Less than solubility product
Much greater than solubility product
Always infinite
Always negative
Medium · Level 4View options
Because more ordered and pure crystals form
Because solvent is destroyed
Because pressure becomes zero
Because solute always becomes gas
Medium · Level 4View options
Solubility changes rapidly with temperature
Solubility does not change with temperature
Solute is always a gas
Solvent is absent
Medium · Level 4View options
Difficult
Very easy
Always complete
Not dependent on pressure
Medium · Level 4View options
Particle size may change rate of dissolving but not final solubility
Small particles always make final solubility infinite
Large particles never dissolve
Particle size alone decides solubility
Medium · Level 4View options
Rate of dissolving may increase but equilibrium solubility does not change
Final solubility always doubles
Solute is destroyed
Mass of solvent becomes zero
Medium · Level 4View options
When temperature and solvent are the same
When temperature and solvent are different
When gases have become solids
When pressure has no meaning
Medium · Level 4View options
Gas solubility decreases
Gas solubility increases
Liquid becomes insoluble
Effect of pressure disappears
Medium · Level 4View options
On carefully cooling a hot saturated solution
Starting with no solute
On completely removing solvent
By always keeping pressure zero
Medium · Level 4View options
Extra solute may separate out
Solution will always remain unsaturated
Solvent will disappear
Pressure will necessarily increase
Medium · Level 4View options
Pressure should decrease gradually so dissolved gas comes out in a controlled way
Solubility of gases in blood should suddenly increase
Water temperature should immediately become very high
The mass of gases should disappear
Medium · Level 4View options
Endothermic process
Exothermic process
Only pressure-controlled process
Completely insoluble process
Question 1MediumLevel 4
Which of two solids will be easier to separate by crystallisation?
Correct answer: A
Crystallization relies on a substantial difference between solubility in hot and cold solvent. A solid whose solubility changes greatly can be dissolved in a small amount of hot solvent and then recovered efficiently on cooling. If solubility hardly changes, little material separates. Complete insolubility also prevents preparation of a suitable solution for crystallization.
If two gases dissolve in the same liquid at the same pressure and the first gas is more soluble, what can be said about its Henry’s law constant?
Correct answer: A
For the convention p = K_H x, the dissolved mole fraction is x = p/K_H. At the same pressure, a gas with a larger x, meaning greater solubility, must have a smaller K_H. Thus, if the first gas is more soluble than the second under identical conditions, its Henry’s law constant is lower. The constants need not be zero, and they are directly related to solubility in this comparison.
If some solvent evaporates from a saturated solution while temperature remains the same, what may happen?
Correct answer: A
A saturated solution contains the maximum amount of solute that can remain dissolved in its present amount of solvent at that temperature. When solvent evaporates, the solvent capacity decreases while the dissolved solute initially remains, so the solution becomes supersaturated temporarily. Excess solute may then crystallize or precipitate until the remaining solution is saturated again.
If the solubilities of two solids change very differently with temperature, which method can be useful for their separation?
Correct answer: A
Fractional crystallisation separates solids by exploiting their different solubilities, especially the different changes in solubility with temperature. A mixture is dissolved, often hot, and then cooled so that the component whose solubility falls more sharply crystallises preferentially. Filtration alone cannot separate two substances that are both dissolved, and boiling into gas is not the general principle here.
Why are divers advised to come up slowly while returning to the surface?
Correct answer: A
At depth, elevated pressure allows more nitrogen and other gases to dissolve in body fluids. A rapid ascent produces a sudden pressure decrease, reducing gas solubility and allowing bubbles to form in tissues and blood. A slow ascent, often with planned decompression stops, gives gases time to leave gradually and reduces the risk of decompression sickness.
In which situation can dissolution of an ionic solid in water be more favourable?
Correct answer: A
Dissolution of an ionic solid requires separation of its ions from the crystal lattice and stabilization of those ions by water molecules. Because water is polar, its molecules can surround and hydrate cations and anions. Strong ion-water attraction favours this step, although the final solubility also depends on lattice energy and the overall energy balance.
What is the effect of the common-ion effect on the solubility of a sparingly soluble salt?
Correct answer: A
Consider a sparingly soluble salt AB in equilibrium with its ions: AB(s) ⇌ A⁺ + B⁻. Adding a soluble electrolyte that supplies A⁺ or B⁻ increases the concentration of that ion. The equilibrium shifts left to oppose the increase, causing more solid AB to remain and reducing its dissolution. Therefore, the common-ion effect decreases solubility, although other complexation or acid–base reactions can modify a specific case.
If one ion of a sparingly soluble salt is added externally to its solution, what will happen?
Correct answer: A
Adding one of the ions already produced by the salt increases a product concentration in its dissolution equilibrium. For example, adding A+ to AB(s) ⇌ A+ + B− causes the equilibrium to shift left according to Le Chatelier's principle. Some dissolved ions combine to form solid salt, so the equilibrium solubility decreases rather than increases.
What is the importance of solubility product for a sparingly soluble salt?
Correct answer: A
The solubility product, Ksp, is the equilibrium constant for dissolution of a sparingly soluble ionic solid, with each ion concentration raised to its stoichiometric power. It describes the ion concentrations in a saturated solution and helps predict precipitation by comparing the ionic product Q with Ksp. It is not a colour, pressure, or container-size measurement.
If the ionic product becomes greater than the solubility product, what is likely to happen?
Correct answer: A
For a sparingly soluble salt, the ionic product Q is calculated from the current ion concentrations, with powers given by the balanced dissolution equation. If Q exceeds K_sp, the solution contains more dissolved ions than equilibrium permits. The excess ions combine to form solid until the ion product falls to K_sp. Thus precipitation occurs. The solution does not become entirely gaseous, and ions do not all disappear.
If ionic product is less than solubility product, what will be the state of the solution?
Correct answer: A
When Q, the actual ionic product, is smaller than Ksp, the ion concentrations have not reached the equilibrium limit for precipitation. The solution is therefore unsaturated with respect to the solid and can dissolve more salt, if solid is available. Some ions may still be present; Q < Ksp does not mean that the solution contains no ions.
If the ionic product is equal to the solubility product, what type of solution is present?
Correct answer: A
The ionic product Q compares the present ion concentrations with the equilibrium limit represented by K_sp. When Q = K_sp, the solution is at equilibrium with the solid phase and contains the maximum equilibrium amount of dissolved salt. It is therefore saturated. If Q is less than K_sp, the solution is unsaturated; if Q is greater than K_sp, precipitation is expected. Hence option A is correct.
Among two salts of similar type, the one with higher solubility product will be what?
Correct answer: A
For salts with the same ion ratio and comparable conditions, a larger Ksp indicates a larger equilibrium product of ion concentrations. This generally corresponds to a greater molar solubility, so that salt is more soluble. The qualification “similar type” matters: for salts with different stoichiometry, Ksp values cannot be compared directly without calculating solubility from the correct expression.
To prevent precipitation in a solution, ionic product should be kept in which range?
Correct answer: A
For a salt, precipitation is expected when the actual ionic product Q becomes greater than Ksp. Keeping Q below Ksp keeps the solution unsaturated with respect to that solid, so precipitation is not thermodynamically required. At Q = Ksp the solution is just saturated; keeping Q much greater would favour precipitation rather than prevent it.
Why is slow cooling considered better than rapid cooling in crystallisation?
Correct answer: A
Slow cooling lowers solubility gradually and gives solute particles enough time to arrange into a regular crystal lattice. Many impurities remain in the mother liquor instead of being trapped inside rapidly growing crystals. Rapid cooling can cause sudden nucleation, producing numerous small crystals that may retain impurities. Thus slow cooling usually improves crystal size and purity.
What does a steep solubility curve of a solid indicate?
Correct answer: A
A solubility curve plots solubility against temperature. Its slope represents the change in solubility per unit change in temperature. A steep slope therefore means that a small temperature change produces a relatively large change in solubility. Such a substance may be suitable for recrystallisation by cooling. A flat curve would indicate little temperature dependence.
If the solubility curve is almost flat, how will purification by crystallisation be?
Correct answer: A
A nearly flat solubility curve means that changing temperature produces only a small change in the amount that dissolves. On cooling a hot solution, little solute will separate, so recovery by crystallisation is poor and purification becomes difficult. Good recrystallisation generally requires a large solubility at high temperature and a much smaller solubility at low temperature.
How is particle size of a solid solute related to its final equilibrium solubility in a liquid?
Correct answer: A
Reducing particle size increases surface area, so solvent can contact more solid and the material usually dissolves faster. However, once equilibrium is reached at a specified temperature and pressure, the final solubility is governed mainly by the chemical nature of the solute and solvent. Particle size does not make solubility infinite; it affects kinetics rather than the equilibrium amount.
What is the effect of stirring on dissolution of a solid?
Correct answer: A
Stirring continually brings fresh solvent into contact with the solid surface and removes the concentration boundary layer around dissolving particles. This usually increases the rate at which the system approaches equilibrium. It does not change the equilibrium solubility at fixed temperature and pressure, so the final maximum dissolved amount is not automatically doubled.
When is it simple to compare gas solubility using only Henry's law constant?
Correct answer: A
Henry's law constant is not universal; it depends on temperature, the gas, and the solvent. If the same solvent and temperature are used, comparing constants for different gases is straightforward at the same pressure: the lower K_H generally corresponds to greater solubility. If temperature or solvent changes, the constants themselves may change, so a direct comparison can be misleading.
What is the practical meaning of an increase in Henry's law constant with temperature?
Correct answer: A
Using p = K_H x, at a fixed gas pressure a larger K_H produces a smaller dissolved mole fraction x. Therefore, if K_H increases as temperature rises, the gas becomes less soluble at that higher temperature. This agrees with the usual behaviour of gases, whose dissolution in liquids is generally less favourable on heating. Pressure still affects solubility; its effect does not disappear.
Under which condition can a solution become supersaturated?
Correct answer: A
A hot saturated solution can contain more dissolved solute than is normally stable at a lower temperature. If it is cooled carefully without shaking or introducing impurities, crystallisation may not start immediately, allowing the excess solute to remain temporarily dissolved. This metastable state is supersaturation. Removing all solvent cannot produce a solution, and zero pressure is not the defining condition.
What may happen when a supersaturated solution is slightly disturbed?
Correct answer: A
A supersaturated solution contains more dissolved solute than the stable equilibrium amount at its current temperature. It is metastable, so a small disturbance can provide nucleation sites or initiate crystal growth. The excess solute then separates as crystals, bringing the solution closer to saturation. The disturbance does not destroy the solvent or necessarily increase pressure; crystallisation is the expected change.
Why are divers advised to return to the surface slowly?
Correct answer: A
At depth, increased pressure causes more gases, especially nitrogen, to dissolve in body fluids. A rapid ascent produces a rapid pressure drop, reducing gas solubility and allowing bubbles to form in tissues or blood. A slow ascent gives the body time to remove the gas gradually. The advice is therefore about controlled decompression, not temperature or loss of mass.
If the solubility of a solid increases with temperature, what type of process is its dissolution generally considered?
Correct answer: A
If heating allows more of a solid to dissolve, heat acts like a reactant that favours the dissolution equilibrium. This observation is therefore generally associated with an endothermic dissolution process. It is a useful qualitative inference, although the exact behaviour of a substance depends on its enthalpy and other solution effects. Pressure control or insolubility does not explain the temperature trend.
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