An element X of atomic mass 25.0 exists as X)_(4) in benzene to the extent of 100%. When 10.30g of saturated solution of X in benzene is added to 20.0 g of benzene, the depression in freezing point of the resulting solution is 0.51 K. If K_(f) for benzene is "5.1 K kg mol"^(-1), the solubility of X in 100 g of benzene will be

An element X of atomic mass 25.0 exists as X)_(4) in benzene to the ex

1.	An element X of atomic mass 25.0 exists as X)_(4) in benzene to the extent of 100%. When 10.30g of saturated solution of X in benzene is added to 20.0 g of benzene, the depression in freezing point of the resulting solution is 0.51 K. If K_(f) for benzene is "5.1 K kg mol"^(-1), the solubility of X in 100 g of benzene will be
Answer» 3.0 g 2.7 g 0.30 g 0.27 g Solution :Suppose saturated solution of X in BENZENE contains w g of X (present as `X_(4)`). Hence, benzene present `=(10.30-w)g` `therefore"Total benzene present"=20+(10.30-w)` `=(30.30-w)g` `DeltaT_(F)=(1000K_(f)w_(2))/(w_(1)xxM_(2))therefore0.51=(1000xx5.1xxw)/((30.30-w)XX100) ` `""(M_(2)or X_(4)=25xx4=100)` `or 51(30.3-w)=5100 w or 30.3-w=100 w` Thus, 10.30 g of saturated solution contains 0.3 g of X and 10.0 g of benzene. `therefore` Solubility of X in 100 g of benzene = 3.0 g