During winter season, in cold countries ethylene glycol is added to water in the car radiators. The amount of ethylene glycol to be added to 2kg of water to prevent freezing at - 3⁰c is (K_f of water = 1.85K K_gmol^-1)

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What mass of ethylene glycol (molar mass 62.0 g mol−1) must be added to 5.50 kg of water to lower the freezing point of water from 00C to -10.00C?(K_fforwater=1.86Kkgmol^{-1}$).

DTf=kf×molality⇒0−(−10)=1.86×w625.5⇒5.5×10=1.86100×w62⇒55=3100×w⇒w=55×1003=18.33=1833gm =1.833kg

What mass of ethylene glycol (molar mass 62.0 g $m o l^{- 1}$ ) must be added to 5.50 kg of water to lower the freezing point of water from $0^{0}$ C to - ${10.0}^{0}$ $C ? ($ K_f $f o r w a t e r = 1.86 K k g$ mol^{-1}$).

$D T_{f} = k_{f} \times m o l a l i t y$
$\Rightarrow 0 - (- 10) = 1.86 \times \frac{\frac{w}{62}}{5.5}$
$\Rightarrow 5.5 \times 10 = \frac{1.86}{100} \times \frac{w}{62}$
$\Rightarrow 55 = \frac{3}{100} \times w$
$\Rightarrow w = \frac{55 \times 100}{3}$
$= 18.33 = 1833 g m$
$= 1.833 k g$