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

Explanation

Molar mass of `CH_(3)CH_(2)CHClCOOH`
= 15 + 14 + 13 + 35.5 + 45 `= 122.5 g mol^(-1)`
Number of moles of 10 g of `CH_(3)CH_(2)CHClCOOH`
`n_(2) = (10)/(122.5) mol`
`= 8.16 xx 10^(-2) mol`
`therefore` Molality of the solution (m)
`= (W_(2) xx 1000)/(M_(2) xx W_(1)) = (n_(2) xx 1000)/(W_(1))`
`= (8.16 xx 10^(-2) mol)/(50g) xx 1000 g kg^(-1)`
`= 0.3264 mol kg^(-1)`
or `C = 0.3264 mol kg^(-1)`
If `alpha` is the degree of dissociation for `CH_(3)CH_(2)CHClCOOH`, then
`CH_(3)CH_(2)CHClCOOH hArr CH_(3)CH_(2)CHClCOO^(-) + H^(+)`
`{:("Initial conc.","C mol L"^(-1),0,0,),("At equilibrium",C(1- alpha),C alpha,C alpha,):}`
`therefore K_(a) = (C alpha. C alpha)/(C(1-alpha)) ~~ C alpha^(2)`
or `alpha = sqrt((K_(a))/(C)) = sqrt((1.4 xx 10^(-3))/(0.3264)) = 0.065`
`CH_(3)CH_(2)CHClCOOH hArr CH_(3)CH_(2)CHClCOO^(-) + H^(+)`
`{:("Initial moles",1,0,0,),("Moles at equilibrium",(1-alpha),alpha,alpha,):}`
Total moles `= 1 - alpha + alpha + alpha = 1 + alpha`
`i = (1 + alpha)/(1) = 1 + alpha = 1 + 0.065 = 1.065`
Therefore, `Delta T_(f) = iK_(f)m`
`= 1.065 xx 1.86 xx 0.3264 = 0.65^(@)C`