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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 $0^{o}$ C to - $10^{o}$ C ? ( $K_{f}$ for water $= 1.86$ K Kg $m o l^{- 1}$ )

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Solution

$Δ T_{f} = K_{f} \times m$
$m = \frac{W_{B}}{M_{B}} \times \frac{1000}{W_{A}}$
$10 = 1.86 \times \frac{W_{B}}{62} \times \frac{1000}{5500}$
$W_{B} = 1833.3 g m$

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