# 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 ${K_b}$ and ${K_f}$ is :a.) ${K_b}$= 0.5 ${K_f}$b.) ${K_b}$= 2 ${K_f}$c.) ${K_b}$= 1.5 ${K_f}$d.) ${K_b}$= ${K_f}$

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Hint: 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-
$\Delta {T_b}$= ${K_b} \times m$
While depression in freezing point describes the lowering of the freezing point of the compound. Its formula is given as-
$\Delta {T_f}$= ${K_f} \times m$
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 ${K_b}$and ${K_f}$
We have the formula to calculate elevation boiling point as-
$\Delta {T_b}$= ${K_b} \times m$
Where $\Delta {T_b}$ is the elevation in boiling point
${K_b}$is the boiling point elevation constant
‘m’ is the molality of the solution
Further, we have the expression for depression in freezing point as-
$\Delta {T_f}$= ${K_f} \times m$
Where $\Delta {T_f}$is the depression in freezing point
${K_f}$ 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,
$\dfrac{{\Delta {T_b}}}{{\Delta {T_f}}}$= $\dfrac{{m \times {K_b}}}{{m \times {K_f}}}$
$\dfrac{2}{2}$= $\dfrac{{1 \times {K_b}}}{{2 \times {K_f}}}$
${K_b}$= 2${K_f}$

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.