Q94P - Chemistry: Molecular Nature Of Matter And Change

Question

Rank the following aqueous solutions in order of increasing

(a) osmotic pressure; (b) boiling point; (c) freezing point;

(d) vapor pressure at 50^oC:

(I) 0.100 m NaNO₃ (II) 0.100 m glucose (III) 0.100 m CaCl₂

Step-by-Step Solution

Answer

 

1. Osmotic pressure: $0.100 m \mathrm{glucose}<0.100 m {\mathrm{NaNO}}_3<0.100 m {\mathrm{CaCl}}_2.$
2. Boiling point: $0.100 m \mathrm{glucose}<0.100 m {\mathrm{NaNO}}_3<0.100 m {\mathrm{CaCl}}_2.$
3. Freezing point: $0.100 \mathrm{m} {\mathrm{CaCl}}_2<0.100 \mathrm{m} {\mathrm{NaNO}}_3<0.100 \mathrm{m} \mathrm{glucose}.$
4. Vapor pressure: $0.100 \mathrm{m} {\mathrm{CaCl}}_2<0.100 \mathrm{m} {\mathrm{NaNO}}_3<0.100 \mathrm{m} \mathrm{glucose}.$

Colligative properties

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.

Calculate molality of particles of each compound

• The  $NaN{O}_3$ will give 2 ions in the solution (sodium and nitrate ions); 2 particles for each molecule of  $NaN{O}_3$.
• A 0.100 m $NaN{O}_3$  will have a 0.200 m molality of ions.
• Glucose is not an ionic compound, so it will not dissociate in other particles and each molecule will dissolve as one particle.
• A 0.100 m glucose will have a 0.100 m molality of glucose.
• The $CaC{l}_2$ will give 3 ions in the solution (2 chloride and 1 calcium ion); 3 particles for each molecule of $CaC{l}_2$.  
• A 0.100 m of  $CaC{l}_2$ will have a 0.300 m molality of ions.

Increasing order according to the calculated molality of the particles

a. As an osmotic pressure is directly proportional to the molality of a solute, the osmotic pressure will be $\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{ }\mathbf{m}\mathbf{ }\mathbf{glucose}\mathbf{<}\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{ }\mathbf{m}\mathbf{ }{\mathbf{NaNO}}_{\mathbf{3}}\mathbf{<}\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{ }\mathbf{m}\mathbf{ }{\mathbf{CaCl}}_{\mathbf{2}}$.

                     

b. The relationship between a molality and the increase in a boiling point is straightforward. So, the higher a molality, the higher will be a boiling point. Thus, $\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{ }\mathbf{m}\mathbf{ }\mathbf{glucose}\mathbf{<}\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{ }\mathbf{m}\mathbf{ }{\mathbf{NaNO}}_{\mathbf{3}}\mathbf{<}\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{ }\mathbf{m}\mathbf{ }{\mathbf{CaCl}}_{\mathbf{2}}.$

 

c. The relationship between a molality and a freezing point depression is direct. So, the higher a molality, the lower a freezing point. thus, $\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{ }\mathbf{m}\mathbf{ }{\mathbf{CaCl}}_{\mathbf{2}}\mathbf{<}\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{ }\mathbf{m}\mathbf{ }{\mathbf{NaNO}}_{\mathbf{3}}\mathbf{<}\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{ }\mathbf{m}\mathbf{ }\mathbf{glucose}.$

 

d. The lowering of a vapor pressure is directly proportional to a molality. So, the higher a molality, the lower a vapor pressure. Thus, $\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{ }\mathbf{m}\mathbf{ }{\mathbf{CaCl}}_{\mathbf{2}}\mathbf{<}\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{ }\mathbf{m}\mathbf{ }{\mathbf{NaNO}}_{\mathbf{3}}\mathbf{<}\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{ }\mathbf{m}\mathbf{ }\mathbf{glucose}.$