The molal elevation constant of water is 0.51. The boiling point of 0.1 molal aqueous NaCl solution is nearly:

100.1°C

The correct answer is option 2. 100.1°C.

To determine the boiling point of a solution, we can use the formula for boiling point elevation:

\(\Delta T_b = iK_b \cdot m \)

where:
\(\Delta T_b\) is the change in boiling point,
\(K_b\) is the molal elevation constant of the solvent, and
\(m\) is the molality of the solution.

For \(NaCl\)

In this case, we have a 0.1 molal aqueous NaCl solution. The molality (\(m\)) is 0.1, and the molal elevation constant (\(K_b\)) for water is given as 0.51.

Substituting these values into the formula, we can calculate the change in boiling point (\(\Delta T_b\)) for the solution.

\( \Delta T_b = 2 \times 0.51 \times 0.1 = 0.102 \)

The change in boiling point is 0.051 degrees Celsius.

To find the boiling point of the solution, we need to add this change to the boiling point of pure water, which is 100 degrees Celsius at normal atmospheric pressure.

\( \text{Boiling point of the solution} = 100 + 0.102 = 100.102 \)

Rounded to the nearest tenth, the boiling point of the 0.1 molal NaCl solution is approximately 100.1°C.

Therefore, the correct option is (2) 100.1°C.