2.7 Kg of each of water and acetic acid are mixed. The freezing point of the solution will be -x °C. Consider the acetic acid does not dimerise in water, nor dissociates in water. x = _______ (nearest integer)

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Solution

Q.
2.7 Kg of each of water and acetic acid are mixed. The freezing point of the solution will be $- x$ °C. Consider the acetic acid does not dimerise in water, nor dissociates in water. $x$ = _______ (nearest integer)

[Given: Molar mass of water = 18 g ${mol}^{- 1}$ , acetic acid = 60 g ${mol}^{- 1}$

$K_{f} H_{2} O : 1.86 K k g m o l^{- 1}$

$K_{f} acetic acid : 3.90 K kg {mol}^{- 1}$

$Freezing point: H_{2} O = 273 K, acetic acid = 290 K$ ] [2024]

Ans.
(31)

$Moles of water = \frac{given mass}{molar mass} = \frac{2700}{18} = 150 mol$

$Moles of acetic acid = \frac{given mass}{molar mass} = \frac{2700}{60} = 45 mol$

As water is in excess, it is solvent.

   $Molality (m) = \frac{n_{acetic acid}}{W_{water} (in kg)} = \frac{45}{2.7} m = 16.667 m$

   $Δ T_{f} = m k_{f} (solvent) = m k_{f} (water)$

$= 16.667 \times 1.86 ° C = 31 ° C$

   $T_{f} (water) - T_{f} (solution) = 31 ° C$

   $0 ° C - T_{f} (solution) = 31 ° C$

   $T_{f} (solution) = - 31 ° C$

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