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0.6mL of glacial acetic acid with densit...

0.6mL of glacial acetic acid with density 1.06`gmL^(-1)` is dissolved in 1kg water and the solution froze at `-0.0205^(@)C`. Calculated van't Hoff factor. `(K_(f)` for water is 1.86J jg `mol^(-1))`

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The correct Answer is:
van't Hoff factor=i=1.0396
Volume of glacial acetic acid=V=0.6mL
Density of glacial acetic acid `=p=1.06gmL^(-1)`
`W_(1)=1kg=1000g`
Freezing point of solution `=t_(f)=-0.0205^(@)C`
`=T_(f)=273+(-0.0205)`
`=272.9795K`
`K_(f)=1.86"K kg mol"^(-1)`
Molar mass of acetic acid=`M_2=60"g mol"^(-1)`
van't Hoff factor =i=?
Depriciation in freezing point= `=DeltaT_((ob))=273-272.9795`
`=0.0205K`
`"Density"=("Mass")/("Volume")`
`therefore "Mass"="Density"xx"Volume"`
`therefore W_(2)=1.06xx0.6=0636g`
`DeltaT_((th))=K_(f)xx(Wxx1000)/(W_(1)xxM_(2))=(1.86xx0.636xx1000)/(1000xx60)`
=0.01972K
`i=(DeltaT_(f_((obs))))/(DeltaT_(f_((th))))=(0.0205)/(0.01972)=1.0396`
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