A sample of an unknown covalent organic compound weighing 1.2 grams is dissolved in 50 grams of pure benzene. The resulting solution freezes at 4.92 degrees C. Determine the molecular weight of this compound. Normally the freezing poing of benzene is 5.48 degrees C and the Kf for benzene is 5.12 deg…
The Correct Answer and Explanation is :
The molecular weight of the unknown covalent organic compound is approximately 219.43 g/mol.
Explanation:
To determine the molecular weight of the unknown compound, we can use the concept of freezing point depression, which is a colligative property that depends on the number of solute particles in the solution, not the nature of the solute. The formula for freezing point depression is:
[
\Delta T_f = K_f \times m
]
Where:
- (\Delta T_f) is the change in freezing point (in °C),
- (K_f) is the freezing point depression constant of the solvent (in this case, benzene),
- (m) is the molality of the solution, defined as moles of solute per kilogram of solvent.
Step 1: Calculate the freezing point depression ((\Delta T_f))
The normal freezing point of benzene is 5.48°C, and the freezing point of the solution is 4.92°C. Thus:
[
\Delta T_f = 5.48°C – 4.92°C = 0.56°C
]
Step 2: Calculate the molality (m)
Using the freezing point depression formula and the known (K_f) for benzene (5.12°C·kg/mol), we can calculate the molality of the solution:
[
m = \frac{\Delta T_f}{K_f} = \frac{0.56°C}{5.12°C·kg/mol} \approx 0.109375 \, \text{mol/kg}
]
Step 3: Calculate the moles of solute
Molality is defined as moles of solute per kilogram of solvent. Since we have 50 grams (0.050 kg) of benzene as the solvent:
[
\text{moles of solute} = m \times \text{mass of solvent in kg} = 0.109375 \, \text{mol/kg} \times 0.050 \, \text{kg} \approx 0.00546875 \, \text{mol}
]
Step 4: Calculate the molecular weight of the solute
Finally, the molecular weight of the solute is given by:
[
M = \frac{\text{mass of solute}}{\text{moles of solute}} = \frac{1.2 \, \text{grams}}{0.00546875 \, \text{mol}} \approx 219.43 \, \text{g/mol}
]
Thus, the molecular weight of the unknown compound is approximately 219.43 g/mol.
Let me now generate a diagram of the process for clarity.
Here is a diagram illustrating the steps to calculate the molecular weight of a solute using freezing point depression. It shows the process, including the calculation of freezing point depression, molality, moles of solute, and ultimately the molecular weight.
