0.6mL of glacial acetic acid with density 1.06gmL^(-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))

0.6mL of glacial acetic acid with density 1.06gmL^(-1) is dissolved in

1.	0.6mL of glacial acetic acid with density 1.06gmL^(-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))
Answer» SOLUTION :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`