A 0.5 percent aqueous solution of KCl was found to freeze at 272.76K. Calculate the Van.t Hoff factor and degree of dissociation of the solute at this concentration (K_(f)" for water = 1.86 k.kg.mol"^(-1)). Normal molar mass of KCl = 74.5.

A 0.5 percent aqueous solution of KCl was found to freeze at 272.76K.

1.	A 0.5 percent aqueous solution of KCl was found to freeze at 272.76K. Calculate the Van.t Hoff factor and degree of dissociation of the solute at this concentration (K_(f)" for water = 1.86 k.kg.mol"^(-1)). Normal molar mass of KCl = 74.5.
Answer» Solution :`T^(@)F" of water "=273K` `T_(f)" of the solution "=272.76K` `DeltaT_(f)=T_(f^(@))-T_(f)=+0.24K` `M_(2)=(K_(f)W_(2))/(DeltaT_(f).W_(1))` Observed molecular mass `M_(2)=(1.86xx0.5xx1000)/(100xx0.24)="38.75 g.mol"^(-1)` The colligative property is INVERSELY related to the molar mass. `therefore"Van.t Hoff factor"` `i=("Observed colligative property")/("NORMAL colligative property")` `=("Theoretical molar mass")/("Observed molar mass")` `"Vant Hoff factor (i) "=(74.5)/(38.75)=1.92` `"Degree of dissociation "alpha=(i-1)/(n-1)` `n="2 for KCL"` `therefore alpha=(1.92-1)/(2-1)` `therefore"Degree of dissociation "=0.92`