If 0.5 g of a solute (molar mass 100 g `mol^(-1)`) in 25 g of solvent elevates the boiling point by 1 K, the molal boiling point constant of the solvent is

Options

A. 2
B. 8
C. 5
D. `0.5`

Explanation

To determine the molal boiling point constant of the solvent, we can use the formula for boiling point elevation:

\[ \Delta T_b = K_b \cdot m \]

Where:
- \(\Delta T_b\) = boiling point elevation (in K)
- \(K_b\) = molal boiling point constant of the solvent (in °C kg/mol)
- \(m\) = molality of the solution (in mol/kg)

### Step 1: Calculate the molality of the solution

Molality (m) is defined as the number of moles of solute per kilogram of solvent.

First, we need to calculate the number of moles of the solute:

\[ \text{Number of moles} = \frac{\text{mass of solute}}{\text{molar mass of solute}} = \frac{0.5 \, \text{g}}{100 \, \text{g/mol}} = 0.005 \, \text{mol} \]

Next, we convert the mass of the solvent from grams to kilograms:

\[ \text{Mass of solvent} = 25 \, \text{g} = 0.025 \, \text{kg} \]

Now we can calculate the molality:

\[ m = \frac{\text{Number of moles of solute}}{\text{mass of solvent in kg}} = \frac{0.005 \, \text{mol}}{0.025 \, \text{kg}} = 0.2 \, \text{mol/kg} \]

### Step 2: Use the boiling point elevation formula

We know that the boiling point elevation \(\Delta T_b\) is given as 1 K. Substituting the values into the boiling point elevation formula:

\[ 1 \, \text{K} = K_b \cdot 0.2 \, \text{mol/kg} \]

### Step 3: Solve for \(K_b\)

Rearranging the equation to solve for \(K_b\):

\[ K_b = \frac{1 \, \text{K}}{0.2 \, \text{mol/kg}} = 5 \, \text{K kg/mol} \]

### Final Answer

The molal boiling point constant of the solvent is:

\[ K_b = 5 \, \text{K kg/mol} \]

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