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Question: How much ethyl alcohol must be added to 1.0 L of water so that solution will not freeze at -4°F? ($K...

How much ethyl alcohol must be added to 1.0 L of water so that solution will not freeze at -4°F? (KfK_f = 1.86°C/m)

A

< 20 g

B

< 10.75 g

C

< 494.5 g

D

494.5 g

Answer

(D)

Explanation

Solution

To prevent water from freezing at -4°F (which is -20°C), the freezing point depression must be at least 20°C. Using the formula ΔTf=Kf×m\Delta T_f = K_f \times m, the required molality is m=20∘C1.86∘C/m≈10.7527 mol/kgm = \frac{20^\circ C}{1.86^\circ C/m} \approx 10.7527 \text{ mol/kg}. For 1.0 L (1.0 kg) of water, this requires approximately 10.7527 moles of ethyl alcohol (C₂H₅OH). Given the molar mass of C₂H₅OH (approx. 46.067 g/mol), the mass of ethyl alcohol needed is 10.7527 mol×46.067 g/mol≈495.22 g10.7527 \text{ mol} \times 46.067 \text{ g/mol} \approx 495.22 \text{ g}. Therefore, more than 494.5 g must be added.