Correct Answer - 0.69 K
`CaCI_(2)" dissociates in aqueous solution"`
`CaCI_(2)(s)overset(("aq"))toCa^(2+)("aq")+2CI^(-)("aq")`
`i=3, W_(B)=10 g, W_(A)=200 g= 0.2, K_(b)=0.512" K kg mol"^(-1)`
`M_(B)=110" g mol"^(-1)`
`DeltaT_(b)=ixxK_(b)m=(ixxK_(b)xxW_(B))/(M_(B)xxW_(A))`
`DeltaT_(b)=((3)xx(0.512" K kg mol"^(-1))xx(10g))/((110" g mol"^(-1))xx(0.2 kg ))=0.69 K`