Match the boiling point with `K_(b)` for x,y and z, if molecular weight of x,y and z are same.
`{:(,b.pt.,"K"_("b")),(x,100,0.68),(y,27,0.53),(z,253,0.98):}`

Explanation

Molal elevation constant may be calculated as,
`K_(b_(1000))=(RT_(0)^(2))/(1000L_(V))" "("where", T_(0)="boiling point of pure solvent "L_(V)= "latent heat of vaporization per gram "L_(V) =(DeltaH_(V))/(m_(B)))`
`= (RT_(0)^(2))/(1000(DeltaH_(V))/(m_(B)))`
`=(RT_(0)^(2)m_(B))/(1000DeltaH_(V))" "("here",DeltaH_(V)= "molar latent heat of vaporisation "m_(B)="molar mass of solute")`
`K_(b_(1000))=(RT_(0)m_(B))/(1000DeltaS_(V))` Since, `DeltaS_(V)=(DeltaH_(V))/(T_(0))`
here, `DeltaS_(V)=` entropy of vaporization
By considering `Delta S_(V)` as almost constant, `K_(b) prop T_(0)`
`:. K_(b)(x)=0.68,K_(b)(y)=0.53andK_(b)(z)=0.98`.