Q13.84P - Chemistry: Molecular Nature Of Matter And Change

Question

Question: Is the boiling point of 0.01 m KF (aq) higher or lower than that of 0.01 m glucose(aq)? Explain.

 

Step-by-Step Solution

Answer

Answer

 

The boiling point of 0.01 m KF (aq) will be higher than that of 0.01 m glucose(aq) because KF will dissociate into ions and produce more particles while glucose does not.

 

Step 1: Concept

 

Strong electrolytes in solution dissociates into ions to give more numberddd of particles. So, we include a multiplying factor in the equations for the colligative properties of electrolyte solutions, which is called van’t Hoff factor (i).

To calculate the colligative properties of strong electrolyte solutions, incorporate the van’t Hoff factor into the equation:$\mathbf{\Delta Tb}\mathbf{=}\mathbf{i}\left({\mathbf{K}}_{\mathbf{b}}\mathbf{m}\right)$

The boiling point elevation for is proportional to the molality and I.

 

Step 2: Calculate van’t Hoff factor (i). for KF and glucose.

 

If strong electrolyte solutions behaved ideally, the factor i would be the amount (mol) of particles in solution divided by the amount (mol) of dissolved solute; that is, i would be 2 for NaCl, 3 for$\mathbf{Mg}\mathbf{(}\mathbf{N}{\mathbf{O}}_{\mathbf{3}}{\mathbf{)}}_{\mathbf{2}}$  etc. 

Assuming solutions to be ideal, 

i = 2 for KF and i = 1 for glucose units. (KF dissociates into 2 particles per formula unit)

 

Step 3: Elevation in boiling point

 

i for KF is twice that of glucose and their molality is same. So, the boiling point elevation of $\left(\mathbf{\Delta Tb}\right)$KF should be twice that of glucose. 

Hence, the boiling point of 0.01 m KF (aq) will be higher than that of 0.01 m glucose(aq).