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Depression in Freezing Point

Kf for water ...

Question

$K_{f}$ for water is $1.86 K k g m o l^{- 1}$ . Your automobile radiator holds 1.0 kg of water, how many grams of ethylene glycol $(C_{2} H_{6} O_{2})$ must you add to get the freezing point of the solution lowered to $- {2.8}^{o} C$ ?

72 g m

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93 g m

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39 g m

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27 g m

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Solution

The correct option is B $93 g m$
$Δ T_{f} = i K_{f} \frac{w_{B} \times 1000}{M_{B} \times w_{A}}$
Given that,
$T_{f} (s o l n) = - {2.8}^{0} C; \Rightarrow Δ T_{f} = 2.8 K$
Let, $x$ be the mass ethylene glycol,

Now,

$2.8 = 1 \times 1.86 \times \frac{x \times 1000}{62 \times 1000}$
By solving the above equation, we get
$x = 93.33 g m$

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