Ethylene glycol, $C H_{2} O H \cdot C H_{2} O H$ , the major component of permanent antifreeze, effectively depresses the freezing point of water in an automobile radiator. What minimum weight of ethylene glycol must be mixed in kg with 6 gallons of water to protect it from freezing at $- 24^{\circ} C . (1 gallon = 3.785 lit, K_{f} = 1.86)$ ?

18.15

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18.16

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18.155

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Solution

Depression in freezing point
$Δ T_{f} = K_{f} \times molality$ ..........(1)
$∵ Δ T_{f} = T_{f}^{0} - T_{f}$
$∵ Δ T_{f} = 0 - (- 24) (Freezing Point of H_{2} O = 0^{\circ})$
$Δ T_{f} = 24^{\circ} C$
Given, $K_{f} = 1.86$
On placing all given values in equation (1), we get
$molality = \frac{Δ T_{f}}{K_{f}} = \frac{24}{1.86} = 12.90 moles$
Thus,
$1 kg of H_{2} O$ should contain 12.90 moles of ethylene glycol
We know,
Molar mass of $C H_{2} O H$ . $C H_{2} O H = 62 g/mol$
Therefore, minimum weight of ethylene glycol required $= (12.90 \times 22.7 \times 62) g$
$= 18155 g or 18.155 kg$ .

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