Q13-120CP - Chemistry: Molecular Nature Of Matter And Change

Question

Four 0.50 m aqueous solutions are depicted below. Assume the solutions behave ideally:

1. Which has the highest boiling point?
2. Which has the lowest freezing point?
3. Can you determine which one has the highest osmotic pressure? Explain.

 

Step-by-Step Solution

Answer

1. Solution A has the highest boiling point
2. Solution D has the lowest boiling point.
3. Solution A has highest osmotic pressure.

Step 1: Ideal solution

The best solutions are those that adhere to Raoult's law at all temperatures and concentrations.

A solution in which the interaction between molecules of the components does not differ from the interaction between the molecules of each component is ideal solution.

Step 2: Highest boiling point

(a)

 Boiling point is the temperature at the pressure exerted by surroundings upon a liquid is equalled by the pressure exerted by the vapor of liquid.

The temperature difference between the solution's boiling point and the pure solvent's boiling point is known as the boiling point elevation. The change in boiling point for a 1-mole solution of a nonvolatile molecular solute is the same as the molal boiling-point elevation constant.

Compute it using the formula below.

$\mathit{\Delta }{\mathit{T}}_{\mathbf{b}}\mathbf{=}{\mathit{T}}_{\mathbf{b}}\mathbf{-}{{\mathit{T}}_{\mathbf{b}}}^{\mathbf{o}}$

Where T_b and T_b^* are the boiling points of pure solvent and solution respectively. 

  $\mathit{\Delta }{\mathit{T}}_{\mathbf{b}}\mathbf{=}\mathit{i}\mathbf{(}{\mathit{K}}_{\mathbf{b}}\mathbf{\times }\mathit{m}\mathbf{)}$

Where $\Delta T_b$is the boiling point elevation, K_b is the boiling point elevation constant, m is molality and i is van’t Hoff factor.

Solution (A) has three particles; therefore, it has van’t Hoff factor of 3.

Solution (B) and (C) has 2 particles; therefore, they have van’t Hoff factor of 2.

Solution (D) has one particle; therefore, it has van’t Hoff factor of 1. 

Molality given for solutions is same and boiling point elevation constant value is also constant for all solution. Since, boiling point is directly proportional to the van’t Hoff factor, solution A has the highest boiling point.

 

Step 3: Lowest freezing point

(b)

 The temperature at which a liquid turns into a solid under normal air pressure is known as the freezing point.

Freezing point depression refers to the lowering of freezing point of solvents upon the addition of solutes. It is a colligative property of solution.

It can by calculated by the following formula

 $\mathit{\Delta }{\mathit{T}}_{\mathbf{f}}\mathbf{=}{{\mathit{T}}_{\mathbf{f}}}^{\mathbf{o}}\mathbf{-}{\mathit{T}}_{\mathbf{f}}$

Where T_f⁰ and T_f are the freezing point of pure solvent and solution respectively.

$\mathit{\Delta }{\mathit{T}}_{\mathbf{f}}\mathbf{=}\mathit{i}\mathbf{(}{\mathit{K}}_{\mathbf{f}}\mathbf{\times }\mathit{m}\mathbf{)}$

Where $\Delta T_f$ is the freezing point depression, K_f is the freezing point depression constant, m is molality and I is van’t Hoff factor.

Molality given for solutions is same and freezing point depression constant value is also constant for all solution. Since, freezing point is directly proportional to the van’t Hoff factor, solution D has the lowest boiling point because it has the lowest value of van’t Hoff factor.

Step 3: Osmotic pressure

 Osmotic pressure, which is frequently employed to indicate the concentration of a solution, is the pressure that would need to be applied to a pure solvent to stop it from osmotically transferring into the supplied solution.

Using the formula below, it can be determined. 

n=iMRT

Where i is van’t Hoff factor, M is molarity of the solution, R is gas constant, T is temperature and is osmotic pressure. 

Molarity is related to the term concentration.   Solution A has the highest concentration of ions.

The van't Hoff factor directly relates to osmotic pressure.

Therefore, solution A has the highest osmotic pressure because it has the highest van’t Hoff factor.