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Answer 1 · 2017-02-04T03:12:58

The depression of freezing point of solvent $DeltaT_f$ is related with the molality $m$ of the solute as follows

$color(blue)(DeltaT_f=-i*K_f*m..........[1])$

where

$K_f->"cryoscopic constant "=1.853Kkg"/mol for water"$

$m->"molality of the soute"=0.100$

$i->" van't hoff factor"=1+alpha$

$alpha->"degree of dissociation"=1.32%=0.0132$

Replacing i by $1+alpha$ the equation [1] becomes

$color(green)(DeltaT_f=-(1+alpha)*K_f*m)$

$color(green)(=>DeltaT_f=-(1+1.0132)*1.853*0.1K~~-0.188K)$

So expected freezing point will be

$color(red)("Expected freezing point"=(273-0.188)K=272.812K)$