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.`

Explanation

`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`