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$^{-1}$).
Correct Answer: 15
Solution and Explanation
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 \( 15 \, \text{kg} \).
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