Calculate the depression in the freezing point of water when 10 g of `CH_(3)CH_(2)CHCl COOH` is added to 250 g of water. `K_(a)=1.4xx10^(-3), K_(f)=1.86" K kg mol"^(-1)`.

Explanation

Molar mass of `CH_(3)CH_(2)CHClCOOH`
`=15+14+13+35.5+12+16+16+1`
`= 122.5 g mol^(-1)`
Therefore, No. of moles present in 10 g `CH_(3)CH_(2)CHClCOOH`
`=(10 g)/(122.5 g mol^(-1))`
= 0.0816 mol
It is given that 10 g of `CH_(3)CH_(2)CHClCOOH` is added to 250 g of water.
Therefore, Molality of the solution,
`=(0.0186)/(250)xx1000`
`=32.64 " mol kg"^(-1)`
Let a be the degree of dissociation of `CH_(3)CH_(2)CHClCOOH`.
`CH_(3)CH_(2)CHClCOOH` undergoes dissociation according to the following equation :
`{:("Initial",CH_(3)CH_(2)CHClCOOH,hArr,CH_(3)CHClCOO^(-),+,H^(+)),("conc. At",C,,O,,O),("equilibrium",C(1-alpha),,C alpha,,C alpha):}`
`K_(a)=(C^(2)x^(2))/(C(1-x))`
Since x is very small with respect to 1 so x can be ignored, `1-alpha =1`
Now,
`K_(a)=(cx^(2))/(1)`
`x=((K_(4))/(C ))^(0.5)`
`= sqrt((1.4xx10^(-3))/(0.3264)) (because K_(a)=1.4xx10^(_3))`
= 0.0655
Again,
`{:("Again,",CH_(3)CH_(2)CHClCOOH,hArr,CH_(3)CHClCOO^(-),+,H^(+)),("initial","1 mol",,0,,0),("equilibrium",1-x,,x,,x):}`
i(Vant Hoff factor) `= 1-x+x+x=1+x`
`= 1+0.0655`
= 1.0655
Hence, the depression in the freezing point of water is given as :
`Delta T_(f)=i.K_(f).m`
`=1.0655xx1.86" K kg mol"^(-1)xx0.3264 " mol kg"^(-1)`
= 0.65 K