Q95P - Chemistry: Molecular Nature Of Matter And Change

Question

Rank the following aqueous solutions in order of decreasing

(a) osmotic pressure; (b) boiling point; (c) freezing point; (d) vapor pressure at 298 K:

(I) 0.04 m urea $\left[{\left({\mathrm{NH}}_2\right)}_{2}C=O\right]$

(II) 0.01 m ${\mathbf{AgNO}}_{\mathbf{3}}$

(III) 0.03 m ${\mathbf{CuSO}}_{\mathbf{4}}$

Step-by-Step Solution

Answer

1. Osmotic pressure:  $0.03{\mathrm{mCuSO}}_4>0.04\mathrm{murea}>0.01{\mathrm{mAgNO}}_3.$
2. Boiling point:  $0.03{\mathrm{mCuSO}}_4>0.04\mathrm{murea}>0.01{\mathrm{mAgNO}}_3.$
3. Freezing point:  $0.01{\mathrm{mAgNO}}_3>0.04\mathrm{murea}>0.03{\mathrm{mCuSO}}_4.$
4. Vapor pressure:  $0.01{\mathrm{mAgNO}}_3>0.04\mathrm{murea}>0.03{\mathrm{mCuSO}}_4.$

Definition

The osmotic and vapor pressures, and boiling and freezing points of a solution are all colligative properties, which mean that these properties depend on the number of particles of solutes in a solution. All these properties depend on the molality of the solutes and ionic solutes get dissociated in a solution to give more number of particles. So, we have to convert the molality of each solute to the molality of the particles in a solution.

 

When the molality of a particle is higher at a given condition, the osmotic pressure as well as the boiling point of the particle will be higher; the freezing point and the vapor pressure of the particle will be lower.

Molality of each given compound

• Urea is not an ionic compound, so it will not dissociate in other particles, and each molecule will dissolve as one particle.
• A 0.04 m urea will have a 0.04 m molality of urea.
• The $AgN{O}_3$  will give 2 ions in the solution (silver and nitrate ions); 2 particles for each molecule of  $AgN{O}_3$.
• A 0.01 m of  $AgN{O}_3$ will have 0.02 m molality of ions.
• The $CuS{O}_4$ will give 2 ions in the solution (copper and sulphate ion); 2 particles for each molecule of $CuS{O}_4$.
• A 0.03 m $CuS{O}_4$  will have a 0.06 m molality of ions.

This means that the $CuS{O}_4$  will have highest molality of particles followed by urea, and $AgN{O}_3$  will have the least molality..

The order of the properties based on the molality of the given compounds

a. Osmotic pressure:  $\mathbf{0}\mathbf{.}\mathbf{03}{\mathbf{mCuSO}}_{\mathbf{4}}\mathbf{>}\mathbf{0}\mathbf{.}\mathbf{04}\mathbf{murea}\mathbf{>}\mathbf{0}\mathbf{.}\mathbf{01}{\mathbf{mAgNO}}_{\mathbf{3}}.$

As osmotic pressure is directly proportional to the molality of a solute, the higher the molality of a compound, the higher will be the osmotic pressure.

 

b. Boiling point: $\mathbf{0}\mathbf{.}\mathbf{03}{\mathbf{mCuSO}}_{\mathbf{4}}\mathbf{>}\mathbf{0}\mathbf{.}\mathbf{04}\mathbf{murea}\mathbf{>}\mathbf{0}\mathbf{.}\mathbf{01}{\mathbf{mAgNO}}_{\mathbf{3}}.$ 

The relationship between a molality and a boiling point of a compound is straightforward. So, the higher the molality, the higher the boiling point.

 

c. Freezing point: $\mathbf{0}\mathbf{.}\mathbf{01}{\mathbf{mAgNO}}_{\mathbf{3}}\mathbf{>}\mathbf{0}\mathbf{.}\mathbf{04}\mathbf{murea}\mathbf{>}\mathbf{0}\mathbf{.}\mathbf{03}{\mathbf{mCuSO}}_{\mathbf{4}}.$

A molality has a direct relationship with a freezing point depression. So, the higher the molality, the lower the freezing point.

 

d. Vapor pressure: $\mathbf{0}\mathbf{.}\mathbf{01}{\mathbf{mAgNO}}_{\mathbf{3}}\mathbf{>}\mathbf{0}\mathbf{.}\mathbf{04}\mathbf{murea}\mathbf{>}\mathbf{0}\mathbf{.}\mathbf{03}{\mathbf{mCuSO}}_{\mathbf{4}}.$

The lowering of a vapor pressure is directly proportional to a molality. So, the higher the molality, the lower the vapor pressure.