The boiling point of water at `750mm Hg` is `99.63^(@)C`. How much sucrose is to be added to `500 g` of water such that it boils at `100^(@)C`.
Given `DeltaT_(b)=100-96.63=3.37^(@)`
Mass of water `w_(1)=500g`
Molar mass of water `M_(1)=18″ g “mol^(-1)`
Molar mass of sucrose, `M_(2)=342″ g “mol^(-1)`
To find: Mass of sucrose `w_(2)=?`
Solution we know, `DeltaT_(b)=K_(b)xxm`
`=K_(b)xx(w_(2))/(M_(2))xx(1000)/(w_(1))`
`impliesw_(2)=(M_(2)xxw_(1)xxDeltaT_(b))/(1000xxK_(b))=(342xx500xx3.37)/(1000xx0.52)`
`w_(2)=1108.2g`
`therefore` mass of solute `w_(2)=1.11` kg
`W_(B)=(M_(B)xxDeltaT_(b)xxW_(B))/K_(b)`
Molar mass of sucrose `(C_(12)H_(22)O_(11))(M_(B))=12xx12+22xx11xx16=342″g mol”^(-1)`
mass pf water`(W_(B))=500g -0.5 kg`
Elevatio in b.p. `(DelataT_(b))=100^(@)C-99.63^(@)C=37^(@)C=0.37 K`
Molal elevation constat `(K_(b))=0.52 K kb mol^(-1))`
`W_(B)=((342g mol^(-1))xx(0.37K)xx(0.5 kg))/((0.52 K kg mol^(-1)))=121.7g`