An element X (Atomic mass = 25) exists as X4 is benzene. 51g of saturated solution of X in benzene was added to 50.0 g of pure benzene. The resulting solution showed a depression of freezing point of 0.55 K. Find the solubility of X per 100 g of benzene. (Kf for benzene = 5.5 K kg "mol"^(-1)

An element X (Atomic mass = 25) exists as X4 is benzene. 51g of satura

1.	An element X (Atomic mass = 25) exists as X4 is benzene. 51g of saturated solution of X in benzene was added to 50.0 g of pure benzene. The resulting solution showed a depression of freezing point of 0.55 K. Find the solubility of X per 100 g of benzene. (Kf for benzene = 5.5 K kg "mol"^(-1)
Answer» Solution :LET X g be the mass of element in 51.0 g of saturated solution. Mass of benzene in 51.0 g of saturated solution = 51.0 -x g Total mass of benzene containing x g of solution `=50+51 -x=(101-x)g` `DeltaT_(f) = (1000 K_(f)W_(B))/(M_(B)W_(A)) = (1000 xx 5.5 xx x)/(4 xx 25 xx (101 -x))` =0.55 (given) `rArr x=1.0 g` Hence, solubility `=(W_(B) xx 100)/W_(A) =1/(51-1) xx 100 = 2.0 g`