Muft Shiksha™ एक 100% Free Education Portal है 🇮🇳, जिसका उद्देश्य Class 9–12 के हर विद्यार्थी तक High-Quality Education को पूरी तरह मुफ्त पहुँचाना है। 🇮🇳 हम मानते हैं कि अच्छी शिक्षा किसी student की आर्थिक स्थिति पर निर्भर नहीं होनी चाहिए। 🇮🇳 हर विद्यार्थी को वही Quality Study Material, MCQs, Quizzes, Exam Preparation, Concept-Based Learning और Bilingual Support मिलना चाहिए, जो आमतौर पर महंगी Coaching या Premium Platforms में मिलता है। Muft Shiksha™ 🇮🇳 इसी सोच के साथ बनाया गया है
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
Quiz this set
Up to 18 questions from this page. Select your focus, then start.
18 questions
Choose questions
Hard · Level 1View options
The same as urea
Half that of urea
Nearly twice that of urea
Nearly three times that of urea
Hard · Level 1View options
When the solute is highly volatile
When the solution is very dilute and contains biomolecules
When the solvent has a very low boiling point
When both solute and solvent are ionic
Hard · Level 1View options
Association of solute molecules
Complete dissociation of solute molecules into ions
Complete conversion of solvent molecules into vapour
Solute concentration being independent of temperature
Hard · Level 1View options
It becomes half
It becomes nearly double
It remains unchanged
It becomes nearly zero
Hard · Level 1View options
Benzoic acid in benzene
Glucose in water
Potassium chloride in water
Urea in water
Hard · Level 1View options
The solute is completely dissociated
The solute is associating
The solute gives two ions
The solute has i = 2.5
Hard · Level 1View options
0.372 K
0.744 K
1.116 K
1.488 K
Hard · Level 1View options
Association of solute molecules
Complete dissociation of the solute
Formation of three ions by each solute unit
Increase in the molality of the solution
Hard · Level 1View options
0.05 M
0.10 M
0.20 M
0.25 M
Hard · Level 1View options
0.85 times
1.00 times
1.18 times
1.30 times
Hard · Level 1View options
0.40
0.60
1.40
1.80
Hard · Level 1View options
The solute has dissociated
The solute has associated
The solute has completely evaporated
No solvent is left
Hard · Level 1View options
Because osmotic pressure is easier to measure than the extremely small boiling-point change
Because proteins always become gases
Because boiling-point elevation is zero for proteins
Because osmotic pressure does not depend on temperature
Hard · Level 1View options
It will be equal
It will be half
It will be double
It will be four times
Hard · Level 1View options
The number of particles will be underestimated, giving an incorrect molar mass
The number of particles will be overestimated, so the result will always be correct
The mass of the solvent will be taken as zero
Freezing-point depression will have no relation to particle number
Hard · Level 1View options
Half of the ideal value
Three-fourths of the ideal value
Equal to the ideal value
Twice the ideal value
Hard · Level 1View options
Solution A
Solution B
Both contain the same number of moles
It cannot be determined
Hard · Level 1View options
0.10 m
0.15 m
0.20 m
0.30 m
Question 1HardLevel 1
A solution contains 0.1 mol of NaCl, which dissociates completely into ions. At the same molality, how will the boiling-point elevation of the NaCl solution compare with that of a urea solution?
Correct answer: C
For a dilute solution, boiling-point elevation is given by ΔTb = iKb m. Urea is a non-electrolyte and does not dissociate, so its van’t Hoff factor is approximately 1. Completely dissociated NaCl produces Na+ and Cl−, giving i approximately 2. Since the molality and solvent are the same, the NaCl solution has nearly twice the boiling-point elevation of the urea solution.
In which situation is osmotic pressure more useful than freezing-point depression for determining molar mass?
Correct answer: B
Osmotic pressure is especially useful for determining the molar mass of proteins, polymers, and other biomolecules because it can be measured in very dilute solutions at ordinary temperatures. These substances may decompose, denature, or undergo other changes when heating is required for boiling-point or freezing-point methods. Since π = CRT for an ideal dilute solution, even a small concentration gives a measurable pressure.
In a solution of a non-volatile solute, the observed colligative effect is lower than the value predicted for ideal behaviour. What does this usually indicate?
Correct answer: A
Colligative effects are proportional to the number of independent solute particles. If solute molecules associate, two or more particles combine to form one larger species, reducing the number of particles in solution. The measured effect then becomes smaller than the ideal prediction, and the van’t Hoff factor is less than one. Complete dissociation would instead increase the effect and give i greater than one.
If CH3COOH forms dimers in benzene, what happens to its observed molar mass?
Correct answer: B
In benzene, acetic acid molecules can associate through hydrogen bonding to form dimers. Thus, two solute molecules behave as one particle, reducing the number of independent particles in the solution. The colligative effect becomes smaller than expected for the unassociated solute. When molar mass is calculated from that reduced effect, the observed molar mass is higher, approximately twice the normal molar mass for complete dimerisation.
In which of the following solutions will the molar mass observed from colligative properties be greater than the actual molar mass because of association of solute molecules?
Correct answer: A
Benzoic acid associates through hydrogen bonding in non-polar benzene and commonly forms dimers. This decreases the effective number of solute particles, so the van’t Hoff factor becomes less than one. Since the observed molar mass is related by Mobserved = Mactual/i, a value of i below one makes the observed molar mass greater than the actual value. KCl, in contrast, dissociates.
In a solution, the observed depression in freezing point is 40% of the expected value for a non-dissociated solute. What is the most appropriate conclusion about the solute?
Correct answer: B
For a fixed solvent and molality, the colligative effect is proportional to the van’t Hoff factor. The observed depression is 40% of the non-dissociated value, so the effective factor is i = 0.40, which is less than 1. A value below 1 indicates that solute particles have associated, reducing the number of independent particles. Complete dissociation would instead produce i greater than 1. Hence, option B is correct.
A 0.4 m non-dissociated solution has a freezing-point depression of 0.744 K. What will be the freezing-point depression of a 0.2 m CaCl₂ solution in the same solvent if CaCl₂ dissociates completely?
Correct answer: C
For the first non-dissociated solution, i = 1 and ΔT_f = K_f m. Therefore, K_f = 0.744/0.4 = 1.86 K kg mol⁻¹. Complete dissociation of CaCl₂ produces three ions, so i = 3. For the second solution, ΔT_f = iK_fm = 3 × 1.86 × 0.2 = 1.116 K. Hence, option C is correct.
The elevation in boiling point, ΔT_b, of a 0.1 m solute solution is lower than the value expected for a non-dissociated solute. What is the most probable reason?
Correct answer: A
Boiling-point elevation is expressed as ΔT_b = iK_bm. For a non-dissociated solute, i = 1. If molecules associate, several solute molecules combine to form fewer independent particles, so i becomes less than 1 and the observed elevation decreases. Dissociation or ionisation would increase the number of particles and therefore increase ΔT_b, not lower it.
Assuming complete dissociation, what is the effective molarity of a 0.05 M K₄[Fe(CN)₆] solution?
Correct answer: D
The salt K₄[Fe(CN)₆] completely dissociates as K₄[Fe(CN)₆] → 4K⁺ + [Fe(CN)₆]⁴⁻. Thus, one formula unit produces five independent solute particles. The effective molarity is therefore iC = 5 × 0.05 M = 0.25 M. The complex ion [Fe(CN)₆]⁴⁻ remains one particle and is not counted as seven separate particles.
In a solution, 30% of the solute particles form dimers. The observed molar mass will be how many times the true molar mass?
Correct answer: C
For dimerisation, two original solute particles combine to form one particle, so the number of particles decreases. If the degree of association is α = 0.30, the van’t Hoff factor is i = 1 − α/2 = 1 − 0.15 = 0.85. Since the observed molar mass is related by Mobserved = Mtrue/i, the required ratio is 1/0.85 = 1.176, approximately 1.18. Thus association makes the observed molar mass larger than the true value.
Two molecules associate to form one dimer. If the degree of association is 80%, what is the van’t Hoff factor?
Correct answer: B
For association in which two molecules form one dimer, the van’t Hoff factor is i = 1 − α/2, where α is the fraction associated. With α = 80% = 0.80, i = 1 − 0.80/2 = 1 − 0.40 = 0.60. Association decreases the total number of particles, so the factor must be less than one. Therefore, option B is correct.
A 0.1 m solution shows a depression in freezing point lower than the expected value for a non-dissociated solute. What is the most suitable conclusion?
Correct answer: B
For a non-dissociated solute, the expected van’t Hoff factor is i = 1, so ΔTf = Kf m. If the observed depression is smaller while the solvent and molality are otherwise comparable, the effective particle factor is less than one. This occurs when solute molecules associate to form dimers or larger aggregates. Association reduces the number of independent particles and therefore lowers the colligative effect.
Why is the osmotic-pressure method more useful than boiling-point elevation for finding the molar mass of a 0.01 M protein solution?
Correct answer: A
For a dilute solution, osmotic pressure is π = CRT, and it can remain measurable even when the concentration is very small. The boiling-point elevation is ΔTb = iKb m; in a dilute protein solution, this temperature change is often extremely small and difficult to determine accurately. Proteins are also non-volatile and may decompose on heating, so avoiding boiling is advantageous. Osmotic measurements are therefore preferred for biomolecules.
A 0.1 M solute has a van’t Hoff factor i = 4. At the same temperature, how will its osmotic pressure compare with that of a 0.4 M non-dissociating solute?
Correct answer: A
For a dilute solution, osmotic pressure is given by π = iCRT, where i is the van’t Hoff factor, C is molar concentration, R is the gas constant, and T is absolute temperature. For the first solution, iC = 4 × 0.1 = 0.4 M. For the non-dissociating solute, i = 1, so iC = 1 × 0.4 = 0.4 M. Since temperature is the same, both solutions have equal osmotic pressure and are isotonic.
A student calculates the molar mass from the depression in freezing point by treating calcium chloride as non-dissociating. What is the main error compared with ideal dissociation?
Correct answer: A
Calcium chloride ideally dissociates according to CaCl2 → Ca2+ + 2Cl−, producing three particles from one formula unit. Therefore its ideal van’t Hoff factor is approximately 3. If the student assumes no dissociation, i = 1 is used, so the effective number of solute particles is underestimated. Because ΔTf = iKf m, the calculation ignores the larger particle effect and produces an erroneous molar mass.
In a solution, solute particles normally remain completely unassociated. If half of the solute particles form dimers, how will the observed depression in freezing point compare with the ideal value at the same molality?
Correct answer: B
Let the original number of solute particles be N. If half, N/2, remain single and the other half, N/2, associate in pairs, those associated particles form N/4 dimers. The total effective particles are therefore N/2 + N/4 = 3N/4. Since freezing-point depression is proportional to the effective particle concentration, ΔTf observed is three-fourths of the ideal value. Thus association gives i = 0.75.
Two solutions A and B have the same osmotic pressure at the same temperature. Solution A contains completely dissociated NaCl, while solution B contains a nonelectrolyte. In the same volume, which solution contains fewer actual moles of solute?
Correct answer: A
Osmotic pressure is π = iCRT. For completely dissociated NaCl, i is approximately 2, whereas for a nonelectrolyte i = 1. Equal osmotic pressures at the same temperature require equal values of iC. Therefore, the NaCl concentration must be about half the nonelectrolyte concentration: 2C_A = C_B. Since the volumes are equal, solution A contains fewer actual moles of solute, even though both solutions have the same effective particle concentration.
A solution contains 0.2 mol of a non-volatile solute in 1 kg of water. If 50% of the solute molecules form dimers, what is the effective molality?
Correct answer: B
The original amount is 0.2 mol in 1 kg of solvent. Half, or 0.1 mol, remains as individual molecules. The other 0.1 mol of molecules associates in pairs, producing only 0.05 mol of dimers. Therefore, effective solute particles are 0.10 + 0.05 = 0.15 mol, and effective molality is 0.15/1 = 0.15 m. Association decreases the number of particles.
Google Analytics helps us understand site usage. Google may send limited cookie-free signals before your choice. The Live Visitors widget operates independently of this analytics choice; see the privacy policy for its provider and fallback details. Essential site features work without analytics cookies. You can change your choice later in Privacy choices. Privacy policy