Calculate the boiling point of a solution prepared by adding 15.00 g of NaCl to 250.0 g of water. `(K_(b)`, for water 0.512 K. kg `mol^(-1)`, molar mus of NaCl = 58.44g)

Explanation

`DeltaT_(b)=K_(b)xxm`
`DeltaT_(b)=K_(b)xx(W_(B))/(M_(B))xx(1000)/(W_(A)" in g")`
`T_(b)-T_(b)^(0)= (0.512xx15xx1000)/(58.44) `
= 0.5256 K
`T_(b) = T_(b)^(0)+DeltaT_(B) = 0.53K`
`T_(b)= 373 +0.5256= 373.53K`