Step 1: Understanding the Concept:
The addition of a nonvolatile solute to a solvent causes an elevation in the boiling point of the solvent.
This elevation is a colligative property and is directly proportional to the molal concentration (molality) of the solution.
Step 2: Key Formula or Approach:
The formula for boiling point elevation is \(\Delta \text{T}_\text{b} = \text{K}_\text{b} \times m\), where \(m\) is molality.
Molality \(m = \frac{n}{\text{W}_{\text{kg}}}\), where \(n\) is moles of solute and \(\text{W}_{\text{kg}}\) is the mass of solvent in kg.
Rearranging the formula gives \(n = \frac{\Delta \text{T}_\text{b} \times \text{W}_{\text{kg}}}{\text{K}_\text{b}}\).
Step 3: Detailed Explanation:
Let's list the known variables from the problem description:
Elevation in boiling point, \(\Delta \text{T}_\text{b} = 0.3 \text{ K}\).
Ebullioscopic constant, \(\text{K}_\text{b} = 1.8 \text{ K kg mol}^{-1}\).
Mass of the solvent in kg, \(\text{W}_{\text{kg}} = 0.3 \text{ kg}\).
We need to solve for \(n\), the number of moles of the solute.
Rearranging the formula to isolate \(n\):
\[ n = \frac{\Delta \text{T}_\text{b} \times \text{W}_{\text{kg}}}{\text{K}_\text{b}} \]
Substitute the known values into the rearranged equation:
\[ n = \frac{0.3 \times 0.3}{1.8} \]
Calculate the numerator:
\[ n = \frac{0.09}{1.8} \]
Perform the division:
\[ n = 0.05 \text{ moles} \]
The exact calculated value is \(0.05\).
Comparing this result to the given options, \(0.051\) (Option A) is numerically the closest.
Such slight discrepancies often occur in competitive exams due to approximations or typos in the original test creation.
In a multiple-choice setting, it is standard practice to select the option nearest to the rigorously calculated correct answer.
Step 4: Final Answer:
Based on the calculation yielding \(0.05\), option (A) is the best choice.