On dissolving 19,5 g of CH_2 FCOOHin 500 g of water, a depression of 1^@Cin freczing pointof water is observed. Calculate the van't Hoff factor and dissociation constant of fluoro acetic acid. Given, K_f = 1.86 K kg "mol"^(-1)

On dissolving 19,5 g of CH_2 FCOOHin 500 g of water, a depression of 1

1.	On dissolving 19,5 g of CH_2 FCOOHin 500 g of water, a depression of 1^@Cin freczing pointof water is observed. Calculate the van't Hoff factor and dissociation constant of fluoro acetic acid. Given, K_f = 1.86 K kg "mol"^(-1)
Answer» Solution :Use the equation : `M_2 = (K_f xx w_2 xx 1000)/(Delta T_fxx w_1)` where `M_2`= observed molecular mass of the solute, `K_f` = molal depression constant, `w_2` = mass of the solute, `w_1 `= mass of the solvent, `Delta T_f` = Depression in FREEZING point Substituting the values in the above equation, we have `M_2 = (1.86 xx 19.5 xx 1000)/(1 xx 500) = 72.54` No. of moles of the ACID n = 19.5/76 = 0.2566 van.t HOFF factor, i=Normal MOLAR mass / Abnormal (observed) molar mass ` = (78)/(72.54) = 1.0753` DISSOCIATION constant `underset(C(1 - alpha))(CH_2FCOOH) iff underset(C alpha)(CH_2FCOO^(-)) + underset(C alpha)(H^(+))` `alpha = i - 1 = 1.0753` `K_a = ([CH_2FCOO^(-) ][H^+])/([CH_2FCOOH]) = (C alpha.C alpha)/(C(1 - alpha)) = (C alpha^2)/(1 - alpha) = C alpha^2` (neglecting a compared to 1) ` C = (19.5 xx 1000)/(78 xx 500) = 0.5 M` `K_a = C alpha^2 = 0.5 xx (0.0753)^2 = 2.83 xx 10^(-3)`