10 gram of solute with molecular mass 100 gram `mol^(-1)` is dissolved in 100 gram solvent to show `0.3^(@)C` elevation in boiling point. The value of molal ebuilioscopic constant will be:

Options

A. 10
B. 3
C. `0.3`
D. un predictable

Explanation

To solve the problem, we need to find the molal ebullioscopic constant (K_b) using the given data. Here’s a step-by-step solution:

### Step 1: Understand the formula for elevation in boiling point
The formula for elevation in boiling point is given by:
\[ \Delta T_b = K_b \cdot m \]
where:
- \(\Delta T_b\) = elevation in boiling point
- \(K_b\) = molal ebullioscopic constant
- \(m\) = molality of the solution

### Step 2: Identify the given values
From the problem, we have:
- Mass of solute = 10 grams
- Molecular mass of solute = 100 g/mol
- Mass of solvent = 100 grams
- Elevation in boiling point (\(\Delta T_b\)) = 0.3 °C

### Step 3: Convert mass of solvent to kilograms
Since molality is defined as moles of solute per kilogram of solvent, we need to convert the mass of the solvent from grams to kilograms:
\[ \text{Mass of solvent in kg} = \frac{100 \text{ grams}}{1000} = 0.1 \text{ kg} \]

### Step 4: Calculate the number of moles of solute
To find the number of moles of solute, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass of solute}}{\text{molecular mass}} = \frac{10 \text{ g}}{100 \text{ g/mol}} = 0.1 \text{ moles} \]

### Step 5: Calculate molality (m)
Now, we can calculate the molality (m):
\[ m = \frac{\text{number of moles of solute}}{\text{mass of solvent in kg}} = \frac{0.1 \text{ moles}}{0.1 \text{ kg}} = 1 \text{ mol/kg} \]

### Step 6: Calculate the molal ebullioscopic constant (K_b)
Now that we have \(\Delta T_b\) and \(m\), we can rearrange the elevation in boiling point formula to solve for \(K_b\):
\[ K_b = \frac{\Delta T_b}{m} = \frac{0.3 \text{ °C}}{1 \text{ mol/kg}} = 0.3 \text{ °C kg/mol} \]

### Final Answer
Thus, the value of the molal ebullioscopic constant (K_b) is:
\[ \boxed{0.3 \text{ °C kg/mol}} \]