Freezing point of ether was lowered by `0.6^(@)C` on dissolving 2.0g of phenol (`C_(6)H_(5)OH`, mol. wt. 94) in 100 g of ether. The molal depression constant for ether is `5.12^(@)`. Calculate the molecular mass of phenol in solution and comment on your result.

Explanation

To solve the problem, we will use the formula for freezing point depression:

\[ \Delta T_f = K_f \times m \]

Where:
- \(\Delta T_f\) = depression in freezing point (in °C)
- \(K_f\) = molal depression constant (in °C kg/mol)
- \(m\) = molality of the solution (in mol/kg)

### Step 1: Identify the given values
- \(\Delta T_f = 0.6 \, °C\)
- \(K_f = 5.12 \, °C \, kg/mol\)
- Mass of phenol (solute) = 2.0 g
- Mass of ether (solvent) = 100 g = 0.1 kg

### Step 2: Calculate molality (m)
First, we need to calculate the number of moles of phenol:

\[ \text{Moles of phenol} = \frac{\text{mass of phenol}}{\text{molar mass of phenol}} = \frac{2.0 \, g}{M} \]

Where \(M\) is the molar mass of phenol we want to find.

Now, we can express molality:

\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{\frac{2.0 \, g}{M}}{0.1 \, kg} = \frac{20}{M} \, mol/kg \]

### Step 3: Substitute into the freezing point depression formula
Now we substitute \(m\) into the freezing point depression equation:

\[ 0.6 = 5.12 \times \frac{20}{M} \]

### Step 4: Solve for M
Rearranging the equation to solve for \(M\):

\[ M = 5.12 \times 20 \div 0.6 \]

Calculating this gives:

\[ M = \frac{102.4}{0.6} = 170.67 \, g/mol \]

### Step 5: Comment on the result
The calculated molar mass of phenol in solution is approximately \(170.67 \, g/mol\), which is significantly higher than the actual molar mass of phenol (94 g/mol). This suggests that phenol dimerizes in the ether solution, meaning that two phenol molecules combine to form a dimer, effectively doubling the apparent molar mass in the solution.