The addition of `0.643 g` of a compound to `50 mL` of benzene (density 0.879 g `mL^(-1)`) lowers the freezing point from `5.51` to `5.03^(@)C`. If `K_(f)` for benzene is `5.12`, calculate the molecular weight of the compound.

Explanation

To solve the problem, we need to calculate the molecular weight of the compound using the freezing point depression method. Here’s a step-by-step solution:

### Step 1: Calculate the change in freezing point (ΔTf)
The change in freezing point (ΔTf) can be calculated as follows:
\[ \Delta T_f = T_f^{\text{initial}} - T_f^{\text{final}} = 5.51^\circ C - 5.03^\circ C = 0.48^\circ C \]

### Step 2: Calculate the mass of the solvent (benzene)
Given the density of benzene is 0.879 g/mL, we can calculate the mass of 50 mL of benzene:
\[ \text{Mass of benzene} = \text{Density} \times \text{Volume} = 0.879 \, \text{g/mL} \times 50 \, \text{mL} = 43.95 \, \text{g} \]

### Step 3: Calculate the mass of the solvent after adding the solute
The mass of the solvent (benzene) after adding the solute is:
\[ \text{Mass of solvent} = \text{Mass of benzene} - \text{Mass of solute} = 43.95 \, \text{g} - 0.643 \, \text{g} = 43.307 \, \text{g} \]

### Step 4: Convert the mass of the solvent to kg
To use in the molality formula, convert the mass of the solvent to kilograms:
\[ \text{Mass of solvent in kg} = \frac{43.307 \, \text{g}}{1000} = 0.043307 \, \text{kg} \]

### Step 5: Calculate the molality (m)
Molality (m) is defined as the number of moles of solute per kilogram of solvent. The number of moles of solute can be expressed as:
\[ \text{Moles of solute} = \frac{\text{mass of solute}}{\text{molecular weight of solute (M)}} = \frac{0.643 \, \text{g}}{M} \]
Thus, the molality is:
\[ m = \frac{0.643/M}{0.043307} = \frac{0.643}{0.043307 \times M} \]

### Step 6: Use the freezing point depression formula
The freezing point depression is related to the molality and the freezing point depression constant (Kf) as follows:
\[ \Delta T_f = K_f \cdot m \]
Substituting the values we have:
\[ 0.48 = 5.12 \cdot \frac{0.643}{0.043307 \times M} \]

### Step 7: Rearranging to find M
Rearranging the equation to solve for M:
\[ M = \frac{5.12 \cdot 0.643}{0.48 \cdot 0.043307} \]

### Step 8: Calculate M
Now, we can calculate M:
\[ M = \frac{5.12 \cdot 0.643}{0.48 \cdot 0.043307} \approx \frac{3.28736}{0.02074936} \approx 158.373 \, \text{g/mol} \]

### Final Answer
The molecular weight of the compound is approximately **158.37 g/mol**.

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