0.400 g of an acid HA (mol. mass = 80) was dissolved in 100 g of water. The solution showed a depression of freezing point of 0.12 K. What will be the dissociation constant (in multiple of 10^(-3) ) of the acid at about 0°C? Given K_f (water) = 1.86 K Kg "mol"^(-1)(Assume molarity of solution a molality)

0.400 g of an acid HA (mol. mass = 80) was dissolved in 100 g of water

1.	0.400 g of an acid HA (mol. mass = 80) was dissolved in 100 g of water. The solution showed a depression of freezing point of 0.12 K. What will be the dissociation constant (in multiple of 10^(-3) ) of the acid at about 0°C? Given K_f (water) = 1.86 K Kg "mol"^(-1)(Assume molarity of solution a molality)
Answer» Solution :MOLALITY (m) of solution `=(0.4 xx 1000)/(80 x 100) = 0.05` `DeltaT_(f)` (normal) `=K_(f) xx m = 1.86 xx 0.05 = 0.093` K Van.t Hoff factor, `i=(DeltaT_(f) ("OBSERVED"))/(DeltaT_(f)("normal")) = 0.12/0.093 = 1.290` `HA + H_(2)O `ALPHA =(i-1)/(n-1) = (1.290-1)/(2-1) = 0.29` `K_(a) = (Calpha^(2))/(1-alpha) = (0.05 xx (0.29)^(2))/(1-0.29) = 5.92 xx 10^(-3)`