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In this Class 12 Chemistry topic from Chapter 01: Solutions, students learn how the physical properties of a solution depend on the number of dissolved solute particles rather than their chemical identity. The topic explains lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure. Students also apply colligative-property equations to calculate molar mass, understand dilute solutions, and use the van’t Hoff factor to interpret association or dissociation of solute particles.
TOPIC PRACTICE
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Up to 25 questions from this page. Select your focus, then start.
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Medium · Level 3View options
Allowing the solvent to flow naturally
Applying external pressure greater than osmotic pressure
Only cooling the solution
Evaporating the solute
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Half
The same
Double
Four times
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Glucose < NaCl < CaCl₂
CaCl₂ < NaCl < glucose
NaCl < glucose < CaCl₂
Glucose = NaCl = CaCl₂
Medium · Level 3View options
Dissociation
Association
Complete ionisation
Increase in evaporation
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It depends only on the nature of the solvent and not on the number of solute particles.
At the same temperature, it increases when the number of solute particles in the solution increases.
It is observed only in solutions containing solid solutes.
It is produced by osmosis even without a semipermeable membrane.
Medium · Level 3View options
1.8
2.2
2.6
3.0
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The vapour pressure of the liquid decreases, so solid–liquid equilibrium is reached at a lower temperature.
The solute always forms ice.
The solvent changes colour.
The mass of the solute is destroyed.
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50 g mol⁻¹
100 g mol⁻¹
150 g mol⁻¹
200 g mol⁻¹
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0.13 K kg mol⁻¹
0.26 K kg mol⁻¹
0.52 K kg mol⁻¹
1.04 K kg mol⁻¹
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It becomes half
It remains the same
It becomes double
It becomes zero
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ΔTf will be less than expected
ΔTf will be more than expected
ΔTf will be infinite
ΔTf will always be zero
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Molarity for ΔTb
Molality for ΔTf
Molality for π
Mass percent for relative lowering of vapour pressure
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0.67
1.0
1.5
2.0
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The osmotic pressure of root-cell sap is greater than that of the soil solution
The soil solution is more concentrated than the root cells
The semipermeable membrane is absent
The soil contains only solid salt and no water
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Sodium chloride solution
Glucose solution
Both will be equal
Neither will show elevation
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Calcium chloride solution
Glucose solution
Both will be the same
They cannot be compared
Medium · Level 3View options
π = CRT
π = C/RT
π = CR
π = RT/C
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Less than the ideal value
Equal to the ideal value
Greater than the ideal value
Zero
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The number of solvent molecules at the surface decreases
The solute evaporates rapidly
The temperature of the solvent decreases
Gas forms in the solution
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At higher effective particle concentration and higher temperature
At lower concentration and lower temperature
At lower temperature for the same concentration
Only when the amount of solvent is greater
Medium · Level 3View options
The one with lower molar mass and more particles
The one with higher molar mass
The one with brighter colour
The one that dissolves less in water
Medium · Level 3View options
0.1 M aluminium chloride
0.1 M glucose
0.1 M urea
0.1 M sucrose
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Dissociation of the solute
Association of the solute
Freezing of the solvent
Change in the colour of the solute
Medium · Level 3View options
Association of the solute
Complete dissociation of the solute
Increase in temperature
Change of the container
Medium · Level 3View options
Their effective particle concentrations are equal
Their masses of solute are equal
Their colours are equal
Their solvents must be different
Question 1MediumLevel 3
What is the basic principle of reverse osmosis used to obtain pure water from seawater?
Correct answer: B
In ordinary osmosis, water passes through a semipermeable membrane from the dilute solution toward the concentrated solution. Reverse osmosis reverses this natural direction by applying an external pressure greater than the solution’s osmotic pressure on the concentrated seawater side. Water is forced through the membrane, while most dissolved salts and other impurities are retained.
A solution contains 0.01 mol solute in 1 L and its osmotic pressure is measured at 300 K. If the solute completely dissociates into 2 ions, how will its osmotic pressure compare with that of an ideal non-dissociated solute?
Correct answer: C
For a dilute solution, osmotic pressure is given by π = iCRT. The concentration C and temperature T are the same in both cases, so the comparison depends only on i. A non-dissociated solute has i = 1, whereas complete dissociation into two ions gives i = 2. Therefore, the dissociated solution has twice the osmotic pressure of the ideal non-dissociated solution.
Which option gives the correct order of boiling points for equal-molality solutions, assuming complete dissociation?
Correct answer: A
Boiling-point elevation is ΔTb = iKb m. For equal molality and the same solvent, the solution with the larger van’t Hoff factor has the greater boiling point. Glucose does not dissociate, so i = 1; NaCl completely dissociates into two ions, so i = 2; and CaCl2 gives three ions, so i = 3. Hence the order is glucose < NaCl < CaCl2.
If the molar mass obtained from freezing-point depression is higher than the true value, what may be happening in the solution?
Correct answer: B
The freezing-point depression is proportional to the number of solute particles: ΔTf = iKf m. If solute molecules associate, fewer independent particles are present and i becomes less than 1. The measured depression is therefore smaller than expected. When the ordinary formula is used without correcting for this smaller effect, the calculated molar mass becomes higher than the true molar mass. Thus, association is responsible.
Which of the following statements about osmotic pressure is correct?
Correct answer: B
Osmotic pressure is the minimum external pressure that must be applied to a solution to stop the net flow of solvent through a semipermeable membrane. For dilute solutions, π = iCRT. At fixed temperature and concentration conditions, increasing the effective number of dissolved particles increases iC and therefore increases osmotic pressure. A semipermeable membrane is essential for osmosis, so options A, C, and D are incorrect.
If a 0.1 m BaCl₂ solution has 80% dissociation, what is the value of the van’t Hoff factor i?
Correct answer: C
BaCl₂ dissociates according to BaCl₂ → Ba²⁺ + 2Cl⁻, so one formula unit produces n = 3 particles on complete dissociation. For a degree of dissociation α, the van’t Hoff factor is i = 1 + α(n − 1). Here α = 80/100 = 0.80, so i = 1 + 0.80(3 − 1) = 1 + 1.60 = 2.60. Therefore option C is correct.
Why does the freezing point of a solution decrease when a non-volatile solute is added?
Correct answer: A
Adding a non-volatile solute lowers the vapour pressure and chemical potential of the liquid solvent. The solid solvent, such as ice, is not affected in the same way. Consequently, the condition in which solid solvent and liquid solution are in equilibrium is achieved only at a lower temperature than for the pure solvent. This lowering of the freezing temperature is called freezing-point depression, so option A is correct.
When 5 g of a solute is dissolved in 250 g of solvent, the molality is 0.2 m. What is the molar mass of the solute?
Correct answer: B
Molality is defined as moles of solute per kilogram of solvent. The solvent mass is 250 g = 0.250 kg. Therefore, moles of solute = molality × mass of solvent in kilograms = 0.2 × 0.250 = 0.050 mol. Molar mass = mass of solute / moles of solute = 5 g / 0.050 mol = 100 g mol⁻¹. Hence option B is correct.
If a 0.5 m non-dissociated solute produces a boiling-point elevation of 0.26 K, what is the value of Kb?
Correct answer: C
For boiling-point elevation, ΔTb = iKb m. The solute is non-dissociated, so its van’t Hoff factor is i = 1. Substituting the given values gives 0.26 K = 1 × Kb × 0.5 mol kg⁻¹. Thus Kb = 0.26/0.5 = 0.52 K kg mol⁻¹. Therefore option C is correct. The unit follows from dividing kelvin by molality.
If the mole fraction of solute is doubled and the solution remains dilute, what happens to the relative lowering of vapour pressure?
Correct answer: C
For a dilute solution containing a non-volatile solute, Raoult’s law gives the relative lowering of vapour pressure as Δp/p° = xsolute, where xsolute is the solute mole fraction. Therefore, as long as the solution remains dilute and ideal, doubling xsolute doubles the relative lowering. The change is direct, not inverse or independent.
In a solution, association of solute molecules makes the van’t Hoff factor i less than 1. What will be the effect on ΔTf?
Correct answer: A
The depression in freezing point is given by ΔTf = iKf m. For a fixed solvent, molality, and Kf, the value of ΔTf is directly proportional to the van’t Hoff factor i. Association combines several solute molecules into fewer particles, making i less than one. Consequently, the actual freezing-point depression is smaller than the value calculated for non-associated particles.
Which option gives the correct concentration unit for calculating the stated colligative property?
Correct answer: B
Freezing-point depression is calculated from ΔTf = Kf m, so the concentration must be expressed as molality, moles of solute per kilogram of solvent. Boiling-point elevation also uses molality, but osmotic pressure uses molarity in π = CRT. Relative lowering of vapour pressure is related to mole fraction, not mass percent. Thus option B is the correct pairing.
If the ideal ΔT_f for a 0.1 m solution is 0.186 K, but the actual ΔT_f is 0.279 K, what is the van't Hoff factor (i)?
Correct answer: C
For freezing-point depression, the relation is ΔT_f = iK_fm. The ideal value corresponds to i = 1, so the van't Hoff factor is obtained by dividing the observed depression by the ideal depression: i = 0.279/0.186 = 1.5. Thus, the solution produces 1.5 times the ideal number of effective particles, usually because of partial dissociation of the solute.
Under which condition can plant roots absorb water from the soil easily?
Correct answer: A
Water enters root cells by osmosis through selectively permeable membranes. A root-cell sap with higher osmotic pressure is generally more concentrated and has lower water potential than the surrounding soil solution. Therefore, water moves from the soil into the root. If the soil solution is more concentrated, it may draw water out of the root and cause difficulty in absorption.
For glucose and sodium chloride solutions of the same molality, which will have a greater elevation in boiling point?
Correct answer: A
Boiling-point elevation follows ΔT_b = iK_bm. Glucose is a non-electrolyte and remains mainly as individual molecules, so i is approximately 1. Sodium chloride ideally dissociates into Na+ and Cl−, giving i approximately 2. At equal molality, the sodium chloride solution therefore contains more effective particles and has the greater boiling-point elevation, although real solutions may show some deviation from ideality.
Between 0.1 molal glucose solution and 0.1 molal calcium chloride solution, which will have the lower freezing point, assuming ideal dissociation?
Correct answer: A
Freezing-point depression is given by ΔT_f = iK_fm. Glucose does not dissociate in water, so i is approximately 1. Under the stated ideal assumption, calcium chloride dissociates as CaCl2 → Ca2+ + 2Cl−, giving i = 3. At the same molality, calcium chloride therefore causes three times the ideal particle effect, a larger depression, and consequently the lower actual freezing point.
For a dilute solution, what is the relation between osmotic pressure (π) and molar concentration (C)?
Correct answer: A
For a dilute solution, osmotic pressure follows an equation analogous to the ideal-gas equation. The correct relation is π = CRT, where π is osmotic pressure, C is the molar concentration of the solute particles, R is the gas constant, and T is the absolute temperature in kelvin. The concentration must be molarity, not molality. If the solute dissociates, the more general relation is π = iCRT, where i is the van’t Hoff factor.
If dissociation of a solute in solution is incomplete, how will the actual van’t Hoff factor compare with the ideal value?
Correct answer: A
The ideal van’t Hoff factor assumes that dissociation is complete and that every formula unit produces the maximum possible number of particles. In incomplete dissociation, only some solute units split into ions, while the rest remain undissociated. Therefore, the actual number of particles is smaller than the ideal number, so the observed van’t Hoff factor is less than the ideal value. The associated colligative effect is consequently smaller than the complete-dissociation prediction.
Why is the vapour pressure of a solution containing a non-volatile solute lower than that of the pure solvent?
Correct answer: A
A non-volatile solute does not appreciably enter the vapour phase. When it dissolves in the solvent, some surface positions are occupied by solute particles, so fewer solvent molecules are present at the surface and able to escape. The escaping tendency of the solvent therefore decreases, lowering its partial vapour pressure. For an ideal solution, Raoult’s law expresses this as p₁ = x₁p₁°, where x₁ is less than one.
In which situation will osmotic pressure be higher?
Correct answer: A
For a dilute solution, osmotic pressure is given by π = iCRT. Thus, at comparable conditions, increasing the effective particle concentration iC or increasing the absolute temperature T increases π. A dissociated electrolyte may have a larger i because it produces more particles. Lower concentration or lower temperature produces a smaller osmotic pressure. The amount of solvent alone is not the determining factor; its effect is represented through the concentration of solute particles.
If equal masses of different solutes are dissolved in equal masses of water, which solute will cause the greater freezing-point depression?
Correct answer: A
Freezing-point depression is given by ΔTf = iKf m. With equal masses of solute and equal masses of water, a solute with lower molar mass provides a larger number of moles because n = mass/molar mass. Therefore its molality is greater and, if the effective particle factor is comparable, it produces a larger depression. If the solutes ionize differently, the van’t Hoff factor must also be considered; the decisive principle is the larger effective number of dissolved particles.
Which solution is likely to have the highest osmotic pressure if temperature is the same?
Correct answer: A
For dilute solutions, osmotic pressure is given by π = iCRT. All four solutions have the same concentration and temperature, so the solution with the greatest van't Hoff factor will have the greatest osmotic pressure. Aluminium chloride is an electrolyte and ideally dissociates as AlCl₃ → Al³⁺ + 3Cl⁻, producing four particles, whereas glucose, urea, and sucrose remain essentially single particles.
If the boiling-point elevation of a solution is greater than the value expected for a non-electrolyte, what may be the reason?
Correct answer: A
Boiling-point elevation is given by ΔT_b = iK_bm. If a solute dissociates into ions, the number of effective particles increases and the van’t Hoff factor becomes greater than one. Consequently, the observed elevation is larger than the value calculated by assuming no dissociation. Association would produce the opposite effect.
If the freezing-point depression of a solution is less than the value expected for a non-associated solute, which reason is most probable?
Correct answer: A
The depression in freezing point is given by ΔT_f = iK_fm. Association joins two or more solute molecules into a larger particle, reducing the number of independently moving particles. Thus, the van’t Hoff factor becomes less than one and the observed freezing-point depression is smaller than the ideal expected value. Dissociation would increase the depression.
Two solutions have equal osmotic pressure at the same temperature. If their solutes are different, what is the best conclusion?
Correct answer: A
For a dilute solution, osmotic pressure is expressed as π = iCRT, where iC represents the effective concentration of solute particles. At the same temperature, equal osmotic pressures imply equal values of iC, even when the chemical solutes are different. Equal osmotic pressure does not require equal mass, colour, or solvent identity.
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