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Standard XII

Chemistry

Depression in Freezing Point

What mass of ...

Question

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}$).

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Solution

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

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Q. What mass of ethylene glycol (Molar mass

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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}$).

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