Pure `CS_2` melts at -112°C. 228 grams of propylene glycol crystals is mixed with 500 grams of `CS_2`. If `k_f` of `CS_2 = -3.83 K kg mol^-1` what is the depression in freezing point?

Options

A. 23°C
B. -135°C
C. -20°C
D. -100°C

Answer

To solve the problem, we need to calculate the depression in the freezing point (ΔTf) of carbon disulfide (CS2) when propylene glycol is added. Here’s a step-by-step solution:

### Step 1: Identify the given data
- Mass of propylene glycol (solute), W2 = 228 g
- Mass of carbon disulfide (solvent), W1 = 500 g
- Freezing point depression constant, Kf = -3.83 K kg mol⁻¹
- Molar mass of propylene glycol (C3H8O2) = 76 g/mol

### Step 2: Calculate the moles of solute (propylene glycol)
To find the number of moles of the solute, we use the formula:
\[ \text{Moles of solute} = \frac{\text{mass of solute}}{\text{molar mass of solute}} \]
Substituting the values:
\[ \text{Moles of solute} = \frac{228 \, \text{g}}{76 \, \text{g/mol}} = 3 \, \text{moles} \]

### Step 3: Convert the mass of solvent from grams to kilograms
Since the molality is defined as moles of solute per kilogram of solvent, we convert the mass of CS2:
\[ \text{Mass of solvent} = 500 \, \text{g} = 0.5 \, \text{kg} \]

### Step 4: Calculate the molality of the solution
Molality (m) is calculated using the formula:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
Substituting the values:
\[ m = \frac{3 \, \text{moles}}{0.5 \, \text{kg}} = 6 \, \text{mol/kg} \]

### Step 5: Calculate the depression in freezing point (ΔTf)
The depression in freezing point can be calculated using the formula:
\[ \Delta Tf = Kf \times m \]
Substituting the values:
\[ \Delta Tf = -3.83 \, \text{K kg/mol} \times 6 \, \text{mol/kg} = -22.98 \, \text{K} \approx -23 \, \text{°C} \]

### Conclusion
The depression in freezing point of CS2 when mixed with propylene glycol is approximately -23 °C.

---