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1000gm of sucrose solution in water is c...

1000gm of sucrose solution in water is cooled to `-0.5^(@)C`. How much of ice would be separated out at this temperature, if the solution started to freeze at `-0.38^(@)C`. Express your answer in gram.
`(K_(f)H_(2)O=1.86K "mol"^(-1)kg)`

Text Solution

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Suppose 100 g of the solution contains `w_(2)` g of the solute and `w_(1)` g of the solvent. Then
`w_(1)+w_(2)=100" …(i)"`
`DeltaT_(f)=(1000xxK_(f)xxw_(2))/(w_(1)xxM_(2))`
As the solution starts freezing at `-0.38^(@)C`. Hence, when `DeltaT_(f)=0.38^(@)`
`0.38=(1000xx1.86xxw_(2))/(w_(1)xx342)" or "(w_(2))/(w_(1))=0.07" ...(ii)"`
Solving eqns. (i) and (ii), we get `w_(2)=6.6g, w_(1)=93.40 g`
Now at `-0.5^(@)C`, some water separates out as ice but solute exists as such, i.e., `w_(2)=6.6g`. Let us caluclate water present `(w_(1))`
`0.50=(1000xx1.86xx6.6)/(w_(1)xx342)" or "w_(1)=71.78`
`therefore" Weight of ice separated out "=93.40-71.78=21.62g`
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