In a `0.2` molal aqueous solution of weak acid `HX` (the degree of dissociation `0.3`) the freezing point is (given `K_(f) = 1.85 K molality^(-1)`):

Options

A. `+0.480^(@)C`
B. `-0.480^(@)C`
C. `-0.360^(@)`
D. `+0.360^(@)`

Answer

2

Explanation

We have
`(DeltaT_(f))_("observed")=(K_(f))_("observed")` molality
`" "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=("Total moles after ionization")/("Total moles before ionozation")`
`=(1-alpha+alpha+alpha)/1=1+alpha`
`=1+0.3`
`=1.3`
we also have
`i=(K_(f) ("observed"))/(K_(f)("calculated"))`
or
`K_(f) ("observed")=i K_(f)("calculate")`
`=(1.3)(1.85)`
`=2.405`
Thus
`DeltaT_(f)=(2.405 K kg mol^(-1))(0.28 mol kg^(-1))`
`=0.481 K` or `0.481^(@)C`
Hence
Freezing point of solution =Freezing point of water `-DeltaT_(f)`
`=(0.000^(@)C)- (0.481^(@)C)`
`= -0.481^(@)C`