By which formula the molal elevation constant (Kb) can be calculated

The elevation in boiling point ($\Delta T_b$) is given by:

$$\Delta T_b = K_b \times m$$

where $K_b$ is the molal elevation constant and $m$ is the molality of the solution.

Molality ($m$) is defined as moles of solute per kilogram of solvent.

$$m = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}}$$

If $W_s$ is the mass of the solute, $M_r$ is the molar mass of the solute, and $W_w$ is the mass of the solvent in grams, then:

$$\text{Moles of solute} = \frac{W_s}{M_r}$$ $$\text{Mass of solvent (in kg)} = \frac{W_w}{1000}$$

So,

$$m = \frac{W_s / M_r}{W_w / 1000} = \frac{W_s \times 1000}{M_r \times W_w}$$

$$\Delta T_b = K_b \times \frac{W_s \times 1000}{M_r \times W_w}$$

Rearranging to solve for $K_b$:

$$K_b = \frac{\Delta T_b \times M_r \times W_w}{1000 \times W_s}$$

Comparing this with the given options, option (a) matches this formula.

The final answer is $\boxed{\text{(a) } \frac{M_r \times \Delta T_b \times W_w}{1000 \times W_s}}$.