A `0.5`molal aqueous solution of a weak acid (HX) is 20 per cent ionized.The lowering in freezing point of this solution is
`(K_(f)=1.86 "K/m for water")`

Options

A. `1.12 K`
B. `-0.56 K`
C. `-1.12 K`
D. `0.56 K`

Answer

1

Explanation

We have
`DeltaT_(f)=iK_(f)m`
We are given
`K_(f)=1.86 K kg mol^(-1)`
`m=0.5 mol kg^(-1)`
We need to find van't hoff factor (i):
`" "HX(aq)hArrH^(+)(aq)+X(aq)`
`{:("Moles before ionization",,1 mol,0 mol,0 mol),("Moles after ionization",,(1-alpha) mol,alpha mol,alpha mol):}`
`i=("Moles after ionization")/("Moles before ionization")`
`=((1-alpha)+(alpha)+(alpha))/1`
`i=1+alpha=1+0.2=1.2`
Substituting thses results, we get
`DeltaT_(f)=(1.2)(1.86 K kg mol^(-1))(0.5 mol kg^(-1))`
`=1.116 K`
Note that `DeltaT_(f)` (depression in freezing point at the solvent) is always a positive quantity.