(a) A solution containing `0.5g` of naphithalene in `50g C Cl_(4)` yield a boiling point elevation of `0.4K`, while a solution of `0.6g` of an unknown solute in the same mass of the solvent gives a boiling point elevation of `0.65K`. Find the molar mass of the unknown solute.
(b) The boiling point of a solution of `0.1g` of a substance in `16g` of ether was found to be `0.100^(@)C` higher that of pure ether. What is the molecular mass of the substance. `K_(b)` (ether) `=2.16K kg "mol"^(-1)`

Explanation

To solve the given problem, we will break it down into two parts as specified in the question.

### Part (a)

1. **Given Data**:
- Mass of naphthalene (W1) = 0.5 g
- Molar mass of naphthalene (M1) = 128 g/mol
- Mass of solvent (CCl4) = 50 g
- Boiling point elevation for naphthalene (ΔTb1) = 0.4 K
- Mass of unknown solute (W2) = 0.6 g
- Boiling point elevation for unknown solute (ΔTb2) = 0.65 K

2. **Formula for Boiling Point Elevation**:
The formula for boiling point elevation is given by:
\[ \Delta T_b = K_b \cdot m \]
where \( m \) is the molality of the solution.

3. **Calculate Molality**:
The molality \( m \) can be expressed as:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{\frac{W}{M}}{\frac{50}{1000}} = \frac{W \cdot 1000}{M \cdot 50} \]
where \( W \) is the mass of the solute and \( M \) is the molar mass of the solute.

4. **Setting Up the Ratios**:
Since \( K_b \) is constant for the same solvent, we can set up the ratio of the boiling point elevations:
\[ \frac{\Delta T_{b1}}{\Delta T_{b2}} = \frac{m_1}{m_2} \]
Substituting the expressions for molality:
\[ \frac{0.4}{0.65} = \frac{\frac{0.5}{128}}{\frac{0.6}{M_2}} \]

5. **Cross Multiplying**:
Cross multiplying gives:
\[ 0.4 \cdot \frac{0.6}{M_2} = 0.65 \cdot \frac{0.5}{128} \]

6. **Solving for M2**:
Rearranging the equation to solve for \( M_2 \):
\[ M_2 = \frac{0.4 \cdot 0.6 \cdot 128}{0.65 \cdot 0.5} \]
Now, calculating the values:
\[ M_2 = \frac{0.4 \cdot 0.6 \cdot 128}{0.65 \cdot 0.5} = \frac{30.72}{0.325} \approx 94.5 \text{ g/mol} \]

### Part (b)

1. **Given Data**:
- Mass of solute = 0.1 g
- Mass of solvent (ether) = 16 g
- Boiling point elevation (ΔTb) = 0.1 °C = 0.1 K
- \( K_b \) for ether = 2.16 K kg/mol

2. **Using the Boiling Point Elevation Formula**:
\[ \Delta T_b = K_b \cdot m \]
Rearranging for molality:
\[ m = \frac{\Delta T_b}{K_b} = \frac{0.1}{2.16} \]

3. **Calculating Moles of Solute**:
Using the definition of molality:
\[ m = \frac{\frac{W}{M}}{\frac{16}{1000}} = \frac{W \cdot 1000}{M \cdot 16} \]
Setting the two expressions for molality equal:
\[ \frac{0.1}{2.16} = \frac{0.1 \cdot 1000}{M \cdot 16} \]

4. **Solving for M**:
Rearranging gives:
\[ M = \frac{0.1 \cdot 1000 \cdot 16}{0.1 \cdot 2.16} \]
Simplifying:
\[ M = \frac{1600}{2.16} \approx 740.74 \text{ g/mol} \]

### Final Answers:
- (a) Molar mass of the unknown solute \( M_2 \approx 94.5 \text{ g/mol} \)
- (b) Molecular mass of the substance \( M \approx 740.74 \text{ g/mol} \)