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Chemistry

5: Colligative Properties

समष्टिगत गुणधर्म

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 18 questions from this page. Select your focus, then start.

18 questions

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Hard · Level 1
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  1. The same as urea
  2. Half that of urea
  3. Nearly twice that of urea
  4. Nearly three times that of urea
Hard · Level 1
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  1. When the solute is highly volatile
  2. When the solution is very dilute and contains biomolecules
  3. When the solvent has a very low boiling point
  4. When both solute and solvent are ionic
Hard · Level 1
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  1. Association of solute molecules
  2. Complete dissociation of solute molecules into ions
  3. Complete conversion of solvent molecules into vapour
  4. Solute concentration being independent of temperature
Hard · Level 1
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  1. It becomes half
  2. It becomes nearly double
  3. It remains unchanged
  4. It becomes nearly zero
Hard · Level 1
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  1. Benzoic acid in benzene
  2. Glucose in water
  3. Potassium chloride in water
  4. Urea in water
Hard · Level 1
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  1. The solute is completely dissociated
  2. The solute is associating
  3. The solute gives two ions
  4. The solute has i = 2.5
Hard · Level 1
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  1. 0.372 K
  2. 0.744 K
  3. 1.116 K
  4. 1.488 K
Hard · Level 1
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  1. Association of solute molecules
  2. Complete dissociation of the solute
  3. Formation of three ions by each solute unit
  4. Increase in the molality of the solution
Hard · Level 1
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  1. 0.05 M
  2. 0.10 M
  3. 0.20 M
  4. 0.25 M
Hard · Level 1
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  1. 0.85 times
  2. 1.00 times
  3. 1.18 times
  4. 1.30 times
Hard · Level 1
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  1. 0.40
  2. 0.60
  3. 1.40
  4. 1.80
Hard · Level 1
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  1. The solute has dissociated
  2. The solute has associated
  3. The solute has completely evaporated
  4. No solvent is left
Hard · Level 1
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  1. Because osmotic pressure is easier to measure than the extremely small boiling-point change
  2. Because proteins always become gases
  3. Because boiling-point elevation is zero for proteins
  4. Because osmotic pressure does not depend on temperature
Hard · Level 1
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  1. It will be equal
  2. It will be half
  3. It will be double
  4. It will be four times
Hard · Level 1
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  1. The number of particles will be underestimated, giving an incorrect molar mass
  2. The number of particles will be overestimated, so the result will always be correct
  3. The mass of the solvent will be taken as zero
  4. Freezing-point depression will have no relation to particle number
Hard · Level 1
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  1. Half of the ideal value
  2. Three-fourths of the ideal value
  3. Equal to the ideal value
  4. Twice the ideal value
Hard · Level 1
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  1. Solution A
  2. Solution B
  3. Both contain the same number of moles
  4. It cannot be determined
Hard · Level 1
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  1. 0.10 m
  2. 0.15 m
  3. 0.20 m
  4. 0.30 m

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