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How many grams of methyl alcohol should be added to 10 L tank of water to prevent its freezing at 268 K?
( $K_{f} o f H_{2} O = 1.86 K - k g / m o l$ )

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880.07 g

899.04 g

860.21 g

878.06 g

answer is C.

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Detailed Solution

$∆ T_{f} = i x K_{f} x m$

(273 – 268) = $1.86 \times \frac{x}{32} \times \frac{1}{10}$

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