The freezing-point depression constant of chloroform is 4.70 kg K/mole. When 10.52 g of a non-volatile solute are dissolved in 150.0 g of chloroform, the solution freezes exactly 2.39 °C below the normal freezing-point of chloroform. What is the molecular weight of the solute?.
The Correct Answer and Explanation is:
To find the molecular weight (or molar mass) of the solute, we can use the formula for freezing point depression:
$$
\Delta T_f = K_f \cdot m
$$
Where:
- $\Delta T_f$ = freezing point depression
- $K_f$ = freezing-point depression constant of the solvent
- $m$ = molality of the solution (moles of solute per kg of solvent)
Step 1: Gather the given data
- $K_f = 4.70 \ \text{kg·K/mol}$
- $\Delta T_f = 2.39^\circ C$ (freezing point depression)
- Mass of solute = 10.52 g
- Mass of chloroform (solvent) = 150.0 g = 0.1500 kg
Step 2: Use formula to find molality (m)
$$
m = \frac{\Delta T_f}{K_f} = \frac{2.39}{4.70} \approx 0.5085 \ \text{mol/kg}
$$
Step 3: Use molality to find moles of solute
Since molality = moles of solute / kg of solvent:
$$
\text{moles of solute} = m \times \text{kg of solvent} = 0.5085 \times 0.1500 \approx 0.07628 \ \text{mol}
$$
Step 4: Calculate molecular weight
Molecular weight (M) is:
$$
M = \frac{\text{mass of solute (g)}}{\text{moles of solute (mol)}} = \frac{10.52}{0.07628} \approx 137.9 \ \text{g/mol}
$$
Final Answer:
$$
\boxed{137.9 \ \text{g/mol}}
$$
Explanation (300+ words):
This problem involves colligative properties, which depend on the number of solute particles in a solvent rather than the nature of the solute itself. One such property is freezing point depression, where the addition of a solute lowers the freezing point of the solvent.
The relationship is expressed through the formula:
$$
\Delta T_f = K_f \cdot m
$$
Where $\Delta T_f$ is the change in freezing point, $K_f$ is the freezing point depression constant for the solvent, and $m$ is the molality of the solution. In this scenario, the freezing point of chloroform decreases by 2.39°C, and the depression constant $K_f$ is 4.70 kg·K/mol.
By dividing the change in temperature by the constant, we find the molality to be approximately 0.5085 mol/kg. Molality measures the number of moles of solute per kilogram of solvent, and with 150.0 g (or 0.1500 kg) of chloroform, we calculate the number of moles of solute to be approximately 0.07628 mol.
Next, using the known mass of the solute (10.52 g), we determine its molecular weight by dividing this mass by the number of moles. This gives a final value of 137.9 g/mol.
This technique is widely used in chemistry to identify unknown substances or to confirm the molecular formula of a compound based on its behavior in solution. It assumes ideal behavior and that the solute is non-volatile and does not dissociate.