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Calculate the depression in the freezing point of water when 10 g of CH subscript 3CH subscript 2CHClCOOH is added to 250 g of water. Ka equal to 1 point 4 into 10 to the power –3 Kf equal to 1 point 86 K kg mol to the power of –1

2.32     Calculate the depression in the freezing point of water when 10 \; g of  CH_{3}CH_{2}CHCICOOH  is added to 250 \; g  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 Ka 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

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