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Calculate boiling point of a solution pe...

Calculate boiling point of a solution pepared by dissoving 15.0 g of NaCI in 250.0 g of water (`K_(b)` for water = 0.512 K kg `mol^(-1)`), Molar mass of NaCI=58.44 g `mol^(-1)`).

Text Solution

Verified by Experts

The correct Answer is:
374.05 K
Step I. Calculation of elevantion in b.p. temperature `(DeltaT_(b))`
NaCI dissociates in aqueous solution as :
`NaCI(s)overset(("aq"))toNa^(+)("aq")+CI^(-)("aq")`
`i=2, W_(B)=15.0 g, M_(B)=58.44" g mol"^(-1), W_(A)=250 g=0.25 kg`
`K_(b)=0.152" K kg mol"^(-1)`
`DeltaT_(b)=ixxK_(b)xxm=(ixxK_(b)xxW_(B))/(M_(B)xxW_(A))`
`DeltaT_(b)=(2xx(0.512" K kg mol"^(-1))xx(15.0g))/((58.44"g mol"^(-1))xx(0.25 Kg))=1.05 K`
Step II. Calculation of boilling point of solution `(T_(b))`
`T_(b)=T_(b)^(@)+DeltaT_(b)=(373+1.05)K=374.05 K`
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