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Heat of Formation

83 g of ethyl...

Question

$83 g$ of ethylene glycol dissolved in $625 g$ of water. The freezing point of the solution is ________ $K$ . (Nearest integer)
[Use : Molal freezing point depression constant of water $= 1.86 K k g m o l^{- 1} .$
Freezing point of water $= 273 K,$
Atomic masses $= C : 12.0 u, O : 16.0 u, H : 10 u]$

269

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269.00

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269.0

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Solution

Freezing point depression of a solution is given by,
$△ T_{f} = i \times K_{f} \times m$
where,
$△ T_{f}$ is the depression in freezing point.
$K_{f}$ is molal depression constant.
m is molality of the solution.
i is Van't hoff factor
i for ethylene glycol is 1.

$Δ T_{f} = (T_{o} - T_{s}) = i \times M o l a l i t y \times K_{f}$
$T_{o}$ is freezing point of water
$T_{s}$ is freezing point of solution

$Molality (m) = \frac{\frac{w_{solute}}{M} \times 1000}{W_{solvent}}$
where,
$w$ is weight of solute
$W$ is weight of solvent
$M$ is molar mass of the solute

$273 - T_{s} = 1 \times \frac{83 \times 1000}{62 \times 625} \times 1.86$

$T_{s} = 269 K$

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