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

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 $15 \, \text{kg}$.