`0.1g` of an unknown substance was dissolved in 5g of camphor and it was found that the melting point of camphor was depressed by `5.3^(@)C`. If `K_(f)` is `39.7`, find the weight of 1 mole of the solute.

Answer

To solve the problem, we will use the formula for depression in freezing point, which is given by:

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

where:
- \(\Delta T_f\) = depression in freezing point
- \(K_f\) = cryoscopic constant of the solvent (camphor in this case)
- \(m\) = molality of the solution

### Step 1: Identify the given values
- Mass of the solute (unknown substance) = \(0.1 \, g\)
- Mass of the solvent (camphor) = \(5 \, g\)
- Depression in freezing point, \(\Delta T_f = 5.3 \, ^\circ C\)
- Cryoscopic constant, \(K_f = 39.7 \, ^\circ C \, kg/mol\)

### Step 2: Convert the mass of the solvent to kilograms
Since molality is defined as moles of solute per kilogram of solvent, we need to convert the mass of camphor from grams to kilograms:

\[ \text{Mass of camphor in kg} = \frac{5 \, g}{1000} = 0.005 \, kg \]

### Step 3: Calculate molality using the depression in freezing point formula
Rearranging the formula \(\Delta T_f = K_f \cdot m\) to find molality \(m\):

\[ m = \frac{\Delta T_f}{K_f} \]

Substituting the known values:

\[ m = \frac{5.3 \, ^\circ C}{39.7 \, ^\circ C \, kg/mol} = 0.133 \, mol/kg \]

### Step 4: Calculate the number of moles of solute
Using the definition of molality:

\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]

Rearranging to find the moles of solute:

\[ \text{moles of solute} = m \cdot \text{mass of solvent in kg} \]

Substituting the values:

\[ \text{moles of solute} = 0.133 \, mol/kg \cdot 0.005 \, kg = 0.000665 \, mol \]

### Step 5: Calculate the molar mass of the solute
The molar mass (molecular weight) of the solute can be calculated using the formula:

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

Substituting the known values:

\[ \text{Molar mass} = \frac{0.1 \, g}{0.000665 \, mol} \approx 150.38 \, g/mol \]

### Final Answer
The weight of 1 mole of the solute is approximately \(150.38 \, g/mol\).

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