When 3.00 g of a substance X' is dissolved in 100 g of CCl$$_4$$, it raises the boiling point by 0.60 K. The molar mass of the substance 'X' is _________ g mol$$^{-1}$$. (Nearest integer).
[Given K$$_b$$ for CCl$$_4$$ is 5.0 K kg mol$$^{-1}$$]

We begin with the well-known formula for elevation of boiling point:

$$\Delta T_b = K_b \, m$$

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

Molality ($$m$$) is defined as

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

The moles of solute can be written in terms of its mass ($$w_2$$) and molar mass ($$M_2$$) as

$$\text{moles of solute} = \frac{w_2}{M_2}$$

Therefore,

$$m = \frac{w_2/M_2}{w_1}$$

Here $$w_2 = 3.00 \text{ g}$$ (mass of substance X), $$w_1 = 100 \text{ g} = 0.100 \text{ kg}$$ (mass of solvent CCl$$_4$$), $$K_b = 5.0 \text{ K kg mol}^{-1}$$, and $$\Delta T_b = 0.60 \text{ K}.$$

Substituting $$m = \dfrac{w_2/M_2}{w_1}$$ into the main formula, we have

$$\Delta T_b = K_b \left(\frac{w_2/M_2}{w_1}\right)$$

Putting in the numerical data:

$$0.60 = 5.0 \left(\frac{3.00/M_2}{0.100}\right)$$

First, simplify the fraction inside the brackets:

$$\frac{3.00/M_2}{0.100} = \frac{3.00}{M_2} \times \frac{1}{0.100} = \frac{3.00}{M_2} \times 10 = \frac{30}{M_2}$$

So the equation reduces to

$$0.60 = 5.0 \left(\frac{30}{M_2}\right)$$

Multiply 5.0 by 30:

$$5.0 \times 30 = 150$$

Thus,

$$0.60 = \frac{150}{M_2}$$

Rearrange to solve for $$M_2$$:

$$M_2 = \frac{150}{0.60}$$

Divide 150 by 0.60:

$$M_2 = 250$$

Hence, the molar mass of substance X is 250 g mol−1.

So, the answer is $$250$$.