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2.32 Calculate the depression in the freezing point of water when of is added to of water.

$K_{a}=1.4\times 10^{-3}$ , $K_{f}=1.86\; K\; kg\; mol^{-1}$

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Firstly we will find the Vant's Hoff factor the dissociation of given compound.

So we can write, $K_a = \frac{(Ca\times Ca)}{C(1-a)}$

or $K_a = Ca^2$ $(\because a << 1)$

or $a = \sqrt{\frac{K_a}{C}}$

Putting values of K_a and C in the last result, we get :

$a = 0.0655$

At equilibrium i = 1 - a + a + a = 1 + a = 1.0655

Now we need to find the moles of the given compound CH₃CH₂CHClCOOH.

So, moles =

$\frac{10}{122.5} = 0.0816\ mol$

Thus, molality of the solution :

$= \frac{0.0816\times1000}{250} = 0.3265\ m$

Now we will use :

$\Delta T_f = i\ K_f\ m$

or $= 1.065 \times 1.86 \times 0.3265 = 0.6467\ K$

Posted by

Devendra Khairwa

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