1.

45 g of ethylene glycol (C_(2)H_(6)O_(2)) is mixed with 600 g water. Calculate (a) freezing point depression (b) freezing point of the solution (K_(f)" for water = 1.86 K kg mol"^(-1)," Atomic masses : C = 12, H = 1, O = 16 amu").

Answer»

SOLUTION :Here, we are given `w_(2)=45g, w_(1)=600g, K_(f)="1.86 K kg mol"^(-1)`
`M_(2)"for SOLUTE "C_(2)H_(6)O_(2)=24+6+32="62 g mol"^(-1)`
Substituting these value in the formula, `DeltaT_(f)=(1000K_(f)w_(2))/(w_(1)xxM_(2)),` we get
`DeltaT_(f)=("1000 g kg"^(-1)xx"1.86 K kgmol"^(-1)xx45g)/(600 gxx"62 g mol"^(-1))="2.25 K"`
Freezing POINT of pure water = 273.15 K
`therefore""` Freezing point of the solution `=T_(f)^(@)-DeltaT_(f)=273.15-2.25K=270.9K`


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