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Raoult's Law

Coolant used ...

Question

Coolant used in car radiator is aqueous solution of ethylene glycol. In order to prevent the solution from freezing at $- {0.3}^{0} C$ . How much ethylene glycol must be added to 5 kg of water? ( $K_{f}$ = 1.86 K $K g m o l^{- 1}$ )

50

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Solution

The correct option is A $50$ g
Ethylene glycol is $C_{2} H_{6} O_{2}$ .
Therefore, using depression formula,
$0.3 = 1.86 \times \frac{n_{e t h y l e n e g l y c o l}}{5} \Rightarrow n = \frac{1.5}{1.86}$
Therefore mass of ethylene glycol $= \frac{1.5}{1.86 \times 62} = 50 g r a m s$

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