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Elevation in Boiling Point

If a automobi...

Question

If a automobile radiator holds 1.0 Kg of water; how many grams of ethylene glycol $(C_{2} H_{6} O_{2})$ must be add to get the freezing point of the solution lowered to $- {2.79}^{\circ} C$ ?
[ $K_{f}$ for water is $1.86 K k g m o l^{- 1}$ ]

93 gm

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39 gm

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27 gm

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72 gm

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Solution

The correct option is A 93 gm
$K_{f} = 1.86 K k g m o l^{- 1}$
$Δ T_{f} = 0 - (- 2.79) = {2.79}^{\circ} C$
Mass of solvent = 1.0 Kg
Mass of solute = ?
Molecular mass of solute = 62 g/mol
$Δ T_{f} = K_{f} \times m$
$m = \frac{\frac{weight of solute}{molecular mass of solute}}{mass of solvent in g} \times 1000$
$m = \frac{W / 62}{1000} \times 1000 = W / 62$
$2.79 = 1.86 \times \frac{W}{62} \Rightarrow W = \frac{62 \times 2.79}{1.86} = 93 g m$

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