The elevation in boiling point for 1 molal solution of glucose is 2 K. The depression in freezing point for 2 molal solution of glucose in the same solvent is 2 K. The relation between $$K_b$$ and $$K_f$$ is:

We recall the two colligative‐property formulas:

For elevation of boiling point, the expression is $$\Delta T_b = K_b \, m$$, where $$\Delta T_b$$ is the rise in boiling temperature, $$K_b$$ is the molal elevation constant of the solvent, and $$m$$ is molality.

For depression of freezing point, the expression is $$\Delta T_f = K_f \, m$$, where $$\Delta T_f$$ is the lowering of freezing temperature, $$K_f$$ is the molal depression constant of the solvent, and $$m$$ is again molality.

We are told that a $$1\,$$molal solution of glucose causes an elevation of $$2\text{ K}$$. Substituting these values into the first formula, we have

$$\Delta T_b = K_b \, m \quad\Rightarrow\quad 2 = K_b \times 1.$$

Dividing both sides by $$1$$ gives

$$K_b = 2.$$

Next, for a $$2\,$$molal solution of the same solute the depression in freezing point is given as $$2\text{ K}$$. Using the second formula, we write

$$\Delta T_f = K_f \, m \quad\Rightarrow\quad 2 = K_f \times 2.$$

Dividing both sides by $$2$$ yields

$$K_f = 1.$$

Now we possess both constants: $$K_b = 2$$ and $$K_f = 1$$. Relating them directly, we can write

$$K_b = 2K_f.$$

Hence, the correct answer is Option A.