A 5% solution of `CaCl_(2)` at `0^(@)C` developed 15 atmospheric pressure. Calculate the degree of dissociation. (mol. wt. (M) of `CaCl_(2) = 111` a.m.u., R = 0.0821 L atm. `K^(-1) "mol"^(-1)`)

Answer

`24.28%`

Explanation

Weight of `CaCl_(2)=(5)/(100) xx 1000=50gL^(-1), T=273K`,
`pi=15"atm", M=111`,
`pi=(iw RT)/(M) or 15=(i xx50xx0.0821 xx 273)/(111)`
`:.i=(15xx111)/(50xx0.0821 xx 273)=1.4857`
`underset((1-alpha))(CaCl_(2)) hArr underset(alpha)(Ca^(2+)) +underset(2alpha)(2Cl^(-))`
`n = 1+2=3`
`a=(i-1)/(n-1)=(1.4857-1)/(3-1)=(0.4857)/(2)`
`=0.2428=24.28%`