Calculate the boiling point of a solution prepared by adding 15.00 g of NaCl to 250 g of water . `(K_(b) = 0.512` K kg `mol^(-1)` and molar mass of NaCl = 58.44 g `mol^(-1)`)

Explanation

`Delta T_(b) = (ik_(b) xx 1000 xx w_(2))/(w_(1) xx M_(2))`
NaCl dissociates as :
`NaCl to Na^(+) + Cl^(-1)`
`therefore " " i = 2`
`w_(2) = 15.0 g , w_(1) = 250.0 g , M_(2) =58.44 g mol^(-1)`
`therefore " " k_(b) = 0.512 K kg mol^(-1)`
`therefore " " Delta T_(b) = (2 xx 0.512 xx 1000 xx 15.0)/(250.0 xx 58.44)`
` = 1.05^(@)C`
`therefore` Boiling point of solution ` = 100+1.05 = 101.05^(@)C`