If 1 g of solute (molar mass = `50 g mol^(-1)`) is dissolved in 50 g of solvent and the elevation in boiling point is 1 K. The molar boiling constant of the solvent is

Options

A. 2
B. 3
C. 2.5
D. 5

Explanation

To solve the problem, we need to calculate the molar boiling constant (Kb) of the solvent using the given data. Let's break it down step by step.

### Step 1: Identify the given data
- Mass of solute (m) = 1 g
- Molar mass of solute (M) = 50 g/mol
- Mass of solvent (W) = 50 g
- Elevation in boiling point (ΔTb) = 1 K

### Step 2: Calculate the number of moles of solute
To find the number of moles of solute (n), we use the formula:
\[ n = \frac{m}{M} \]
Substituting the values:
\[ n = \frac{1 \text{ g}}{50 \text{ g/mol}} = 0.02 \text{ mol} \]

### Step 3: Calculate the mass of the solvent in kilograms
The mass of the solvent needs to be converted from grams to kilograms:
\[ W = 50 \text{ g} = \frac{50}{1000} \text{ kg} = 0.05 \text{ kg} \]

### Step 4: Calculate the molality (m)
Molality (m) is defined as the number of moles of solute per kilogram of solvent:
\[ m = \frac{n}{W} \]
Substituting the values:
\[ m = \frac{0.02 \text{ mol}}{0.05 \text{ kg}} = 0.4 \text{ mol/kg} \]

### Step 5: Use the formula for elevation in boiling point
The formula relating the elevation in boiling point to the molar boiling constant is:
\[ \Delta Tb = Kb \times m \]
We need to rearrange this formula to solve for Kb:
\[ Kb = \frac{\Delta Tb}{m} \]

### Step 6: Substitute the values to find Kb
Now, substituting the known values into the equation:
\[ Kb = \frac{1 \text{ K}}{0.4 \text{ mol/kg}} = 2.5 \text{ K kg/mol} \]

### Conclusion
The molar boiling constant of the solvent is:
\[ Kb = 2.5 \text{ K kg/mol} \]