How much ethyl alcohol $C_{2} H_{5} O H$ must be added to $1.00 L$ of water so that the solution will not freeze at $- 4 F^{o}$ ?

211 g

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495 g

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85 g

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46 g

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Solution

The correct option is B $495 g$
$4^{o} F = \frac{5}{9} {(- 4 - 32)}^{o} C = - 20^{o} C$
$Δ T_{f} = 20^{o} C = K_{f} m = ({1.86}^{o} C / m) (m)$
$m = \frac{20^{o} C}{{1.86}^{o} C / m} = 10.70 m = (10.7 m o l) (46.0 g / m o l) = 495 g$

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