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A solution of sucrose (molar mass =324 g...

A solution of sucrose (molar mass =324 g/mol) is prepared by dissolving 68.4 g in 1000 g of water. What is the
freezing point and
boiling point of the solution ? `K_(f)`= water is 1.86 K/m and `K_(b)` =0.52 K/m.

Text Solution

Verified by Experts

The correct Answer is:
f.p=272.63K, b.p. = 373.1 K
`M_(B)=342" g mol"^(-1), W_(B)=68.4 g, W_(A)=1 kg,`
`K_(f)=1.86Km^(-1), K_(b)=0.52Km^(-1),T_(f)=?, T_(b)=?`
`DeltaT_(f)=(K_(f)xxW_(B))/(M_(B)xxW_(A))" "DeltaT_(f)=(K_(b)xxW_(B))/(M_(B)xxW_(A))`
`=((186" K kg mol"^(-1))xx(68.4 g))/((342" g mol"^(-1))xx(1kg))," "DeltaT_(b)=((0.52" K kg mol"^(-1))xx(68.4g))/((342g)xx(1 kg))`
`=0.372 K," " =0.104 K`
`T_(f)=T_(f)^(@)-DeltaT_(f)=273-0.372," "T_(b)=T_(b)^(@)+DeltaT_(b)=373+0.104`
`=272.628 K" "=373.104 K`
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