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 calculate the molecular weight of the compound, we will follow these steps:

### Step 1: Calculate the change in freezing point (ΔTf)
The change in freezing point (ΔTf) is 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 benzene (W1)
To find the mass of benzene, we use the formula:
\[ \text{Mass} = \text{Density} \times \text{Volume} \]
Given that the density of benzene is \(0.879 \, \text{g/mL}\) and the volume is \(50 \, \text{mL}\):
\[ W_1 = 0.879 \, \text{g/mL} \times 50 \, \text{mL} = 43.95 \, \text{g} \]

### Step 3: Use the freezing point depression formula
The freezing point depression formula is given by:
\[ \Delta T_f = K_f \times m \]
where \(m\) is the molality of the solution. The molality (m) can be expressed as:
\[ m = \frac{W_2}{M_2} \times \frac{1000}{W_1} \]
Here, \(W_2\) is the mass of the solute (0.643 g), and \(M_2\) is the molar mass of the compound which we need to find.

### Step 4: Substitute known values into the formula
Substituting the known values into the freezing point depression formula:
\[ 0.48 = 5.12 \times \left(\frac{0.643}{M_2} \times \frac{1000}{43.95}\right) \]

### Step 5: Rearranging to find M2
Rearranging the equation to solve for \(M_2\):
\[ 0.48 = 5.12 \times \frac{0.643 \times 1000}{M_2 \times 43.95} \]
\[ M_2 = \frac{5.12 \times 0.643 \times 1000}{0.48 \times 43.95} \]

### Step 6: Calculate M2
Now, we will perform the calculations:
\[ M_2 = \frac{5.12 \times 0.643 \times 1000}{0.48 \times 43.95} \approx \frac{3296.56}{21.096} \approx 156.06 \, \text{g/mol} \]

### Final Answer
The molecular weight of the compound is approximately:
\[ \boxed{156.06 \, \text{g/mol}} \]