Boiling point of chloroform is 61°C. After addition of 5.0 g of a non-volatile solute to 20 g chloroform boils at 64.63°C. If kb = 3.63 K kg mol-1, what is the molecular weight of the solute?

Options

A. 320 g/mol
B. 100 g/mol
C. 400 g/mol
D. 250 g/mol

Answer

To solve the problem, we need to determine the molecular weight of the non-volatile solute based on the boiling point elevation of chloroform. Here are the steps to find the solution:

### Step 1: Calculate the boiling point elevation (ΔTb)
The boiling point elevation is calculated by subtracting the original boiling point of chloroform from the new boiling point after the solute is added.

\[ \Delta T_b = T_{b, \text{new}} - T_{b, \text{original}} = 64.63°C - 61°C = 3.63°C \]

### Step 2: Use the boiling point elevation formula
The formula for boiling point elevation is given by:

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

where:
- \( k_b \) is the ebullioscopic constant (3.63 K kg mol⁻¹ for chloroform),
- \( m \) is the molality of the solution.

### Step 3: Calculate the molality (m)
Molality is defined as the number of moles of solute per kilogram of solvent. To find molality, we need to rearrange the formula:

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

Substituting the values:

\[ m = \frac{3.63°C}{3.63 \, \text{K kg mol}^{-1}} = 1 \, \text{mol/kg} \]

### Step 4: Calculate the number of moles of solute
Since we have 20 g of chloroform, we convert this to kilograms:

\[ \text{mass of solvent} = 20 \, \text{g} = 0.020 \, \text{kg} \]

Using the definition of molality:

\[ \text{moles of solute} = m \cdot \text{mass of solvent in kg} = 1 \, \text{mol/kg} \cdot 0.020 \, \text{kg} = 0.020 \, \text{mol} \]

### Step 5: Calculate the molecular weight of the solute
Molecular weight (M) can be calculated using the formula:

\[ M = \frac{\text{mass of solute}}{\text{moles of solute}} \]

Substituting the values:

\[ M = \frac{5.0 \, \text{g}}{0.020 \, \text{mol}} = 250 \, \text{g/mol} \]

### Final Answer
The molecular weight of the solute is **250 g/mol**.

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