To `250 mL` of water, `x g` of acetic acid is added. If `11.5%` of acetic acid is dissociated, the depressin in freezing point comes out `0.416`. What will be the value of `x` if `K_(f) ("water")=1.86 K kg^(-1)` and density of water is `0.997 g mL`.

Explanation

Given that: `W_(2)=x g` ,`DeltaT_(f)=0.416`,`alpha=0.115`
`alpha=(i-1)/(n-1)`
So, `i=1.115`
`DeltaT_(f)=ixxmmxxK_(f)`
`DeltaT_(f)=ixx(W_(2)xx1000xxK_(f))/(Mw_(2)xxW_(1))`
`0.416=1.115xx(x xx 1000 xx1.86)/(60xx249.25)` `[W_(2)= x g]`
x=`(0.416xx60xx249.25)/(1.115xx1000xx1.86)=3 g`