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Lowering of vapour pressure, Deltap, ele...

Lowering of vapour pressure, `Deltap`, elevation in boiling point, `DeltaT`, and depression in freezing point, `DeltaT_f` of a solvent for the same molar concentration of each of the three solutes : `(A) sugar, (B) NaCl, and (C) BaCl_2`, follow the sequence :

A

`Deltap : A lt B lt C`

B

`DeltaT_b : C gt B gt A`

C

`DeltaT_f : Alt Blt C`

D

all of these

Text Solution

Verified by Experts

The correct Answer is:
D
If solute dissociates in the solution, e.g., `NaCl, MgCl_2, BaCl_2`, etc., the number of particles in solution will be more than that actually dissolved and value of colligative properties will be higher.
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Knowledge Check

  • Lowering of vapour pressure: Deltap, elevation in boiling point: Delta T_(b), and depression in freezing point : Delta T_(f) of a solvent for the same molar concentration of each of the three solutes: A. sugar, B. NaCl and C. BaCl_(2 ) follow the sequence

    A
    `Delta p : A lt B lt C`
    B
    `Delta T _(b) : C gt B gt A`
    C
    `Delta T _(f) : A lt B lt C`
    D
    All of these

  • Consider lowering of vapour pressure (Deltap) , elevation in boiling point (Delta T_(b)) and depression in freezing point (Delta T_(f)) of a solvent for the same molar concentration of each of the following three solutes : 1. BaCl_(2) 2. NaCl 3. MgCl_(2) Which of the following is/are the correct sequence ?

    A
    `Delta p : 3 lt 2 lt 1`
    B
    `Delta T_(b) : 1 gt 2 gt 3`
    C
    `Delta T_(f) : 3 lt 2 lt 1`
    D
    None of these

  • If for the sucrose solution elevation in boiling point is 0.1^(@)C then what will be the boiling point of NaCl solution for same molal concentration

    A
    `0.1^(@)C`
    B
    `0.2^(@)C`
    C
    `0.08^(@)C`
    D
    `0.01^(@)C`

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    For the same solution the elevation in boiling point has higher values than depression in freezing point.

    If for a sucrose solution elevation in boiling point is 0.1^(@)C then what will be the boiling point of NaCl solution for same molal concentration

    If for a sucrose, elevation in boiling point is 1.0^(@)C , then what will be the boiling point of NaCl solution for same molal concentration?

    A system of greater disorder of molecules is more probable. The disorder of molecules is reflected by the entropy of the system. A liquid vapourizes to form a more disordered gas. When a solute is present, there is additional contribution to the entropy of the liquid due to increased randomness. As the entropy of solution is higher than that of pure liquid, there is weaker tendency to form the gas. Thus, a solute (non-volatile) lowers the vapour pressure of a liquid, and hence a higher boiling point of the solution. Similarly, the greater randomness of the solution opposes the tendercy to freeze. In consequence, a lower temperature must be reached for achieving the equilibrium between the solid (frozen solvent) and the solution. The elevation in boiling point (DeltaT_(b)) and depression in freezing point (DeltaT_(f)) of a solution are the colligative properties which depend only on the concentration of particles of the solute and not their identity. For dilute solutions, (DeltaT_(b)) and (DeltaT_(f)) are proportional to the molarity of the solute in the solution. To aqueous solution of Nal , increasing amounts of solid Hgl_(2) is added. The vapour pressure of the solution

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