Question
Question: Elevation in the boiling point for 1 molal solution of glucose is 2 K. The depression in the freezin...
Elevation in the boiling point for 1 molal solution of glucose is 2 K. The depression in the freezing point of 2 molal solution of glucose in the same solvent is 2 K. The relation between Kb and Kf is :
a.) Kb= 0.5 Kf
b.) Kb= 2 Kf
c.) Kb= 1.5 Kf
d.) Kb= Kf
Solution
The elevation in boiling point describes that when a compound is added to another compound, its boiling point rises and the added compound is impure. Its formula can be given as-
ΔTb= Kb×m
While depression in freezing point describes the lowering of the freezing point of the compound. Its formula is given as-
ΔTf= Kf×m
On filling the values and by dividing, we can get the answer.
Complete step by step answer :
Let us start by writing what is given to us and what we need to find.
Thus, Given :
Molality of the solution A = 1 molal
Elevation in the boiling point of A = 2 K
Molality of the solution B = 1 molal
Depression in the freezing point of B = 2 K
To find : Relation between Kband Kf
We have the formula to calculate elevation boiling point as-
ΔTb= Kb×m
Where ΔTb is the elevation in boiling point
Kbis the boiling point elevation constant
‘m’ is the molality of the solution
Further, we have the expression for depression in freezing point as-
ΔTf= Kf×m
Where ΔTfis the depression in freezing point
Kf is the cryoscopic constant or molal depression constant
‘m’ is the molality of solution.
On dividing the elevation in boiling point by depression in freezing point
We have,
ΔTfΔTb= m×Kfm×Kb
22= 2×Kf1×Kb
Kb= 2Kf
So, the option b.) is the correct answer.
Note: It must be noted that the elevation in boiling point and depression in freezing point are colligative properties. These depend on the number of solute particles present in the solution.