To observe an elevation of boiling point of `0.05^(@)C`, the amount of a solute (molecular weight = 100) to be added to 100 g of water `(K_(b) = 0.5)` is

Options

A. 2 g
B. 0.05 g
C. 1 g
D. 0.75 g

Explanation

To solve the problem, we will use the formula for the elevation of boiling point, which is given by:

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

Where:
- \(\Delta T_b\) is the elevation in boiling point,
- \(K_b\) is the ebullioscopic constant of the solvent (water in this case),
- \(m\) is the molality of the solution.

### Step 1: Identify the given values
- \(\Delta T_b = 0.05^\circ C\)
- \(K_b = 0.5^\circ C \cdot kg/mol\)
- Molar mass of solute = 100 g/mol
- Mass of solvent (water) = 100 g = 0.1 kg (since molality is expressed in kg of solvent)

### Step 2: Rearrange the formula to find molality
We can rearrange the formula to solve for molality \(m\):

\[ m = \frac{\Delta T_b}{K_b} \]

### Step 3: Substitute the values into the formula
Now, substituting the known values into the equation:

\[ m = \frac{0.05}{0.5} = 0.1 \, mol/kg \]

### Step 4: Calculate the number of moles of solute
Molality is defined as the number of moles of solute per kg of solvent. Therefore, we can express it as:

\[ m = \frac{n}{mass \, of \, solvent \, (kg)} \]

Where \(n\) is the number of moles of solute. Rearranging gives:

\[ n = m \cdot mass \, of \, solvent \, (kg) \]

Substituting the values:

\[ n = 0.1 \, mol/kg \cdot 0.1 \, kg = 0.01 \, moles \]

### Step 5: Calculate the mass of the solute
Now, we can find the mass of the solute using the number of moles and the molar mass:

\[ mass \, of \, solute = n \cdot molar \, mass \]

Substituting the values:

\[ mass \, of \, solute = 0.01 \, moles \cdot 100 \, g/mol = 1 \, g \]

### Final Answer
The amount of solute to be added to 100 g of water to observe an elevation of boiling point of \(0.05^\circ C\) is **1 g**.

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