Question

Mass of ethylene glycol (antifreeze) to be added to 18.6 kg of water to protect the freezing point at $-24^\circ C$ is ____ kg (Molar mass in g mol$^{-1}$ for ethylene glycol = 62, $K_f$ of water = 1.86 K kg mol$^{-1}$).

Answer 1

Correct Answer: 15

To solve the problem of determining the mass of ethylene glycol needed to protect water from freezing at $-24^\circ C$, we'll use the concept of freezing point depression, which is given by the formula: 
$\Delta T_f = K_f \cdot m$, where $\Delta T_f$ is the change in freezing point, $K_f$ is the cryoscopic constant, and $m$ is the molality of the solution.

1. Determine the Freezing Point Depression ($\Delta T_f$):
The normal freezing point of water is $0^\circ C$. The desired freezing point is $-24^\circ C$, so:
$\Delta T_f = 0 - (-24) = 24^\circ C$.

2. Calculate Molality (m):
$m = \frac{\Delta T_f}{K_f} = \frac{24}{1.86} = 12.903$ mol kg$^{-1}$.

3. Determine the Moles of Solute Needed:
Molality is defined as moles of solute per kilogram of solvent, so:
moles of ethylene glycol = $12.903 \times 18.6$ kg = 239.9998 mol.

4. Convert Moles to Mass:
Using the molar mass of ethylene glycol (62 g mol$^{-1}$):
Mass of ethylene glycol = $239.9998$ mol $\times 62$ g mol$^{-1}$ = 14,879.9876 g = 14.88 kg

5. Verify the Result Within the Range:
The computed mass of ethylene glycol is 14.88 kg, which falls within the given range of 15±0.15 kg (14.85 kg to 15.15 kg).

Thus, the mass of ethylene glycol to be added is 14.88 kg.

Answer 2

Step-by-step Calculation
The depression in freezing point (\( \Delta T_f \)) is given by:
\[\Delta T_f = K_f \times m\]where:
\( \Delta T_f = 24^\circ\text{C} \) (since the freezing point is to be lowered to \( -24^\circ\text{C} \))
\( K_f = 1.86 \, \text{K kg mol}^{-1} \) (cryoscopic constant of water)
\( m \) is the molality of the solution.
Rearranging the formula to find molality:
\[m = \frac{\Delta T_f}{K_f} = \frac{24}{1.86} \approx 12.903 \, \text{mol kg}^{-1}\]
Calculating the Mass of Ethylene Glycol:
Molality (\( m \)) is defined as:
\[m = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}}\]
Let \( n \) be the number of moles of ethylene glycol. Therefore:
\[n = m \times \text{mass of solvent} = 12.903 \times 18.6 \approx 240.9958 \, \text{mol}\]
The mass of ethylene glycol is given by:
\[\text{Mass of ethylene glycol} = n \times \text{molar mass of ethylene glycol}\]
Substituting the known values:
\[\text{Mass of ethylene glycol} = 240.9958 \times 62 \approx 14,941.74 \, \text{g} \approx 15 \, \text{kg}\]
Conclusion:The mass of ethylene glycol required is approximately \( 15 \, \text{kg} \).