Correct Answer - `P_(s)=23.78mm`
`CaCl_(2)hArrCa^(2+)+2Cl^(-)`
`{:(1,0,0),(1alpha,alpha,2alpha):}`
`therefore` Total particles `=1+2alpha`
`therefore i=1+2xx0.9=2.8`
Also molality of `CaCl_(2)` solution `=(2xx1000)/(111xx100)=0.180M`
`therefore pi=C.S.Txxi`
`=0.180xx0.0821xx300xx2.8`
`=12.41` atm
Since, Molariy `=` Molality `=0.180`
`DeltaT_(b)=K_(b)xx"molality"xxi`
`=0.51xx0.180xx2.8`
`=0.257`
`therefore` b.pt. of solution `=100+0.257=100.257^(@)C`
`DeltaT_(f)=K_(f)x` molality `xi=1.86xx0.180xx2.8=0.938`
`therefore` Freezing point of solution `=-0.938^(@)C`.
`(P^(@)-P_(s))/(P_(@))=(nxxi)/(nxxi+N)xxi`
or `(P^(0)-P_(s))/(i.P^(0))=(n)/(n+N)`
or `(i.P^(0))/(P^(0)-P_(s))=(n+N)/(n)=1+(N)/(n)`
`(i.P^(0)-P^(0)+P_(s))/(P^(0)-P_(s))=(1000)/(0.180xx18`
`therefore P_(s)=23.78mm`