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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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एक से कम
एक के बराबर
एक से अधिक
शून्य
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आयनीकरण
संघटन
कणों की संख्या में वृद्धि
वाष्प-दाब में पूर्ण वृद्धि
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अविद्युत अपघट्य विलेय का विलयन
ऐसे विलेय का विलयन जो दो आयन देता है
दोनों विलयनों में समान वृद्धि होगी
संघटन करने वाले विलेय का विलयन
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बहुत तनु विलयन में भी परासरण दाब मापा जा सकता है
बड़े अणु हमेशा आयन बनाते हैं
इस विधि में तापमान की आवश्यकता नहीं होती
यह विधि केवल रंग मापती है
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कोशिका के अंदर
कोशिका से बाहर
कोई शुद्ध प्रवाह नहीं होगा
केवल विलेय दोनों दिशाओं में जाएगा
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विलायक का कोई शुद्ध प्रवाह नहीं होगा
विलायक केवल पहले विलयन में जाएगा
विलायक केवल दूसरे विलयन में जाएगा
दोनों विलयन तुरंत उबलेंगे
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क्योंकि यह संबंध निरपेक्ष तापमान पर आधारित है
क्योंकि सेल्सियस पैमाने में सांद्रता नहीं होती
क्योंकि केल्विन में विलेय गायब हो जाता है
क्योंकि रंग केवल केल्विन में मापा जाता है
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The lowering decreases
The lowering increases
The lowering becomes zero
There is no relation
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Because the vapour pressure of the solution decreases
Because the mass of the solution becomes zero
Because the solute forms vapour and increases pressure
Because the colour of the solvent changes
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Because the solute interferes with the orderly freezing of solvent molecules
Because the solute always releases heat
Because the solvent changes into a metal
Because no particles remain in the solution
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0.45 K
0.90 K
1.80 K
3.60 K
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0.10 K
0.25 K
0.50 K
1.00 K
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They are determined only by the nature of the solute
They depend on the effective number of solute particles
They depend only on colour
They depend only on the shape of the container
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1
2
3
0.5
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They will be the same
The first will be greater
The second will be greater
Both will be zero
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The one in which ionisation occurs
The one in which ionisation does not occur
Both will always be the same
Neither solution will show any effect
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Identify the applicable property and formula
Write the colour of the solute
Draw the shape of the container
Add all the numbers together
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Their molar concentrations are equal
Their colours are equal
Their solutes are identical
Their densities are equal
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Because solvent molecules become heavier
Because the mole fraction and escaping tendency of solvent molecules decrease
Because the solute itself is highly volatile
Because the temperature automatically decreases
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C₆H₁₂O₆
NaCl
CaCl₂
Al₂(SO₄)₃
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Complete association has occurred
Partial dissociation has occurred
No dissociation has occurred
The solute has become the solvent
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Glucose
NaCl
CaCl₂
Urea
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0.002
0.02
0.20
2.0
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0.186 K
0.372 K
0.744 K
1.116 K
Medium · Level 2View options
Association of solute molecules
Dissociation of solute
Freezing of the solvent
Stopping evaporation of the solute
Question 1MediumLevel 2
If the observed depression in freezing point is greater than the value expected for a non-electrolyte because of ionisation, what is the van’t Hoff factor, i?
Correct answer: C
The van’t Hoff factor i represents the ratio of the actual number of solute particles in solution to the number expected without association or dissociation. Ionisation splits a solute into two or more ions, increasing the effective particle count. Since ΔT_f = iK_fm, a larger-than-expected depression means i > 1.
If the observed elevation in boiling point, ΔT_b, is less than the value expected for a non-electrolyte, which process is most likely responsible?
Correct answer: B
For a dilute solution, boiling-point elevation is given by ΔT_b = iK_bm. Association causes two or more solute particles to combine into larger units, so the number of independent particles decreases and i becomes less than one. Consequently, the observed ΔT_b is smaller than the value calculated for an undissociated solute.
Which solution will show the greatest elevation in boiling point at the same molality, assuming complete dissociation and comparable solvent conditions?
Correct answer: B
Boiling-point elevation follows ΔT_b = iK_bm. At the same molality and with the same solvent, K_b and m are fixed, so the elevation depends on the van’t Hoff factor i. A non-electrolyte gives i approximately 1, whereas a solute that completely forms two ions gives i approximately 2. Therefore option B produces the greater elevation.
Why is the osmotic-pressure method particularly useful for determining the molar mass of large molecules?
Correct answer: A
Large molecules such as proteins and polymers may decompose or undergo structural changes when their solutions are heated, so boiling-point or freezing-point methods can be unsuitable. Osmotic pressure can be measured accurately at or near room temperature using a very dilute solution. Since π = CRT, the method gives molar mass without requiring a large amount of solute or heating.
If a cell is placed in a hypertonic solution, in which direction will the net movement of water occur?
Correct answer: B
A hypertonic external solution has a higher effective solute concentration, and therefore a lower water potential, than the cell interior. Across a selectively permeable membrane, water moves by osmosis from the side with higher water potential to the side with lower water potential. Thus water leaves the cell, causing it to shrink or become plasmolysed.
If two solutions are separated by a semipermeable membrane and have equal osmotic pressure at the same temperature, what will happen?
Correct answer: A
Solutions having equal osmotic pressure at the same temperature are isotonic with respect to the membrane. Although individual solvent molecules may cross the semipermeable membrane in both directions, the two opposing rates are equal. Therefore there is no net movement of solvent from one solution to the other, and no volume change due to osmosis occurs.
Why must temperature be expressed in kelvin in the osmotic-pressure equation π = iCRT?
Correct answer: A
The equation π = iCRT is derived from the ideal-solution or gas-like relationship in which temperature is measured from absolute zero. The kelvin scale is an absolute scale, so its zero corresponds to the minimum theoretical thermal energy and ratios of temperature remain physically meaningful. Using Celsius directly would give an incorrect proportionality because 0 °C is not absolute zero.
If the effective number of solute particles in a solution increases, what happens to the lowering of vapour pressure?
Correct answer: B
Lowering of vapour pressure is a colligative property, so it depends on the effective number of solute particles rather than their chemical identity. When more particles are present, fewer solvent molecules can escape into the vapour phase. Consequently, the vapour pressure decreases by a larger amount. Ionisation increases the particle count and therefore can increase the lowering.
Why does the boiling point of a solution become higher than that of the pure solvent when a non-volatile solute is added?
Correct answer: A
A non-volatile solute does not contribute appreciably to the vapour phase, but it reduces the escaping tendency of solvent molecules. Thus, at any given temperature, the solution has a lower vapour pressure than the pure solvent. The solution must be heated to a higher temperature before its vapour pressure equals the external pressure, so its boiling point rises.
Why is the freezing point of a solution lower than that of the pure solvent?
Correct answer: A
Dissolved solute particles disturb the formation of an ordered crystal lattice by solvent molecules. Therefore, the chemical potential of the liquid solvent becomes relatively lower than that of the pure solvent, and a lower temperature is required for solid-liquid equilibrium. This decrease in freezing temperature is called depression of freezing point and is a colligative effect.
If 0.1 mol of a non-electrolyte solute is dissolved in 200 g of solvent and K_f = 1.8 K kg mol⁻¹, what is the depression in freezing point, ΔT_f?
Correct answer: B
First convert the solvent mass into kilograms because molality is moles of solute per kilogram of solvent: 200 g = 0.200 kg. Therefore, m = 0.1/0.200 = 0.5 mol kg⁻¹. For a non-electrolyte, i = 1, so ΔT_f = iK_fm = 1 × 1.8 × 0.5 = 0.90 K. Thus, option B is correct.
If 0.2 mol of a non-electrolyte solute is dissolved in 400 g of solvent and K_b = 0.5 K kg mol⁻¹, what is the elevation in boiling point, ΔT_b?
Correct answer: B
Convert the solvent mass first: 400 g equals 0.400 kg. The molality is therefore m = 0.2/0.400 = 0.5 mol kg⁻¹. Since the solute is a non-electrolyte, the van’t Hoff factor is i = 1. Applying ΔT_b = iK_bm gives ΔT_b = 1 × 0.5 × 0.5 = 0.25 K. Therefore, the correct answer is option B.
Which statement gives the correct understanding of colligative properties?
Correct answer: B
Colligative properties are properties of dilute solutions that depend primarily on the number of dissolved particles relative to the amount of solvent, not on the chemical identity, colour, or shape of the container. Their magnitude is affected by dissociation or association because these processes change the effective particle count. Thus, option B is the correct statement.
If a solute completely dissociates into three ions in solution, what is the ideal van’t Hoff factor, i?
Correct answer: C
The van’t Hoff factor i is the ratio of the actual number of solute particles in solution to the number expected if no dissociation or association occurred. Under ideal complete dissociation, one formula unit producing three ions gives three effective particles. Therefore, i = 3, and option C is correct. This value increases colligative effects threefold compared with an undissociated solute.
If two non-electrolyte solutions have the same molality and the same solvent, how will their ΔT_f values compare?
Correct answer: A
Freezing-point depression is given by ΔT_f = iK_fm. For non-electrolytes under ideal conditions, i = 1. Because both solutions use the same solvent, they have the same K_f, and the stated molalities are equal. Every factor in the equation is therefore identical, so both solutions have the same ΔT_f. Option A is correct.
If two solutions have the same molality but the solute ionises in one of them, which solution will show the greater colligative effect?
Correct answer: A
Colligative properties depend on the total number of dissolved particles, not merely on the original number of solute formula units. Ionisation changes one solute unit into two or more ions, increasing the effective particle concentration. Therefore, at the same molality, the ionising solute produces the greater elevation, depression, or osmotic effect. This increase is represented by the van’t Hoff factor, i, which is greater than one for ideal dissociation.
In a numerical problem on colligative properties, what should be done first to obtain the correct answer?
Correct answer: A
A colligative-properties problem must first be classified correctly. Determine whether it involves relative lowering of vapour pressure, boiling-point elevation, freezing-point depression, or osmotic pressure. Then select the corresponding equation, identify quantities such as molality, mole fraction, concentration, or van’t Hoff factor, convert units consistently, and only afterward substitute values and calculate.
Two solutions have the same osmotic pressure at the same temperature. If both are ideal dilute solutions, which statement is correct?
Correct answer: A
For an ideal dilute solution, osmotic pressure is given by π = CRT, where C is the molar concentration, R is the gas constant, and T is absolute temperature. Because both solutions have the same π and the same T, and R is universal, their molar concentrations must be equal. This does not require identical solutes, colours, or densities; those properties depend on chemical identity and composition.
Why does the vapour pressure of a solvent decrease when a non-volatile solute is added?
Correct answer: B
When a non-volatile solute is dissolved, it contributes essentially no vapour above the solution. It reduces the mole fraction of the solvent and therefore reduces the fraction of surface molecules able to escape into the vapour phase. Raoult’s law expresses this as p₁ = x₁p₁⁰; because x₁ is less than one, the solvent vapour pressure is lower than that of the pure solvent at the same temperature.
Among aqueous solutions of equal molality, assuming complete dissociation, which solute will produce the greatest depression in freezing point?
Correct answer: D
The depression in freezing point is given by ΔT_f = iK_fm. Since all solutions have the same solvent and the same molality, K_f and m are constant, so the largest van’t Hoff factor produces the greatest depression. Glucose gives one particle, NaCl gives two ions, CaCl₂ gives three ions, and Al₂(SO₄)₃ gives five ions on complete dissociation. Therefore option D is correct.
In a solution, the van’t Hoff factor of KCl is 1.8. If ideal complete dissociation gives a value of 2, what does this value indicate?
Correct answer: B
KCl ideally dissociates as KCl → K⁺ + Cl⁻, so one formula unit can produce two particles and the ideal van’t Hoff factor is 2. A measured value of 1.8 lies between the value for no dissociation, approximately 1, and the value for complete dissociation, 2. Thus the particles are not completely separated; the result indicates partial dissociation, possibly with some ion pairing.
Among the following solutions of equal molality in the same solvent, which will have the lowest freezing point if dissociation is complete?
Correct answer: C
Freezing-point depression is calculated using ΔT_f = iK_fm. At equal molality in the same solvent, the solution with the largest van’t Hoff factor has the greatest depression and therefore the lowest actual freezing point. Glucose and urea are non-electrolytes with i approximately 1, NaCl gives about 2 ions, and CaCl₂ gives about 3 ions. Hence CaCl₂ is correct.
For a solution, Δp/p⁰ = 0.02. Assuming a very dilute solution with a non-volatile solute, what is the approximate mole fraction of the solute?
Correct answer: B
For a solution containing a non-volatile solute, Raoult’s law gives the relative lowering of vapour pressure as (p⁰ − p)/p⁰ = x₂, where x₂ is the solute mole fraction. Therefore, if Δp/p⁰ is 0.02, the approximate solute mole fraction is x₂ = 0.02. The dilute-solution assumption supports this simple relation and rules out values such as 0.20 or 2.0.
If a 0.2 m glucose solution has ΔT_f = 0.372 K, what will be the approximate ΔT_f of a 0.2 m NaCl solution in the same solvent, assuming complete dissociation?
Correct answer: C
The relation for freezing-point depression is ΔT_f = iK_fm. Glucose is a non-electrolyte, so i = 1, and its given depression is 0.372 K. At the same molality and in the same solvent, completely dissociated NaCl produces Na⁺ and Cl⁻, so i = 2. Its depression is therefore twice the glucose value: 2 × 0.372 = 0.744 K. Hence option C is correct.
If the boiling point elevation of a solution is greater than the expected value, what could be a possible reason?
Correct answer: B
The elevation in boiling point is given by ΔTb = iKb m. If the expected calculation assumes i = 1 but the solute dissociates into two or more particles, the van’t Hoff factor becomes greater than 1. Consequently, the number of solute particles increases and the observed boiling-point elevation becomes larger than the expected value. Therefore, dissociation is the correct reason.
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