The molecular weight of an organic compound is 58 g `"mol"^(-1)`. What will be the boiling point of a solution containing 48 g of the solute in 1200 g of water?

Explanation

Molecular weight = 58 gm/mol.
Boiling point of solution = ?
48 gm solute in 1200 gm `H_(2)O`
`K_(b)` for water = `0.513^(@)C`
Boiling point of water = `100^(@)C(T_(b))`
`DeltaT_(b)=(1000xxK_(b)xxw)/(MxxW)`
`DeltaT_(b)=(0.513xx1000xx48)/(1200xx58)`
`DeltaT_(b)=0.353^(@)C`
As we know that,
`DeltaT_(b)=T-T_(b)`
`0.353^(@)C=T-100^(@)C`
`100.353^(@)C=T`
`therefore` B.P. of the solution is `100.353^(@)C`.