18g of glucose , `C_6H_(12)O_6` (Molar mass=180g `mol^-1`) is dissolved in 1000g (1kg) of water in a sauce pan . At what temparature will water boil at 1.013 bar? `K_b` for water is 0.52K kg `mol^-1` . Water boils at 373.15K at 1.013bar pressure.

Explanation

Mass of solute = 18 g
Moles of solute = `18/(180)=0.1`
Mass of solvent = 1 kg
`T_(b)=?`
`DeltaT_(b)=K_(b)*m=0.52xx0.1xx1/(1)=0.052`
`DeltaT_(b)=T_(b)-T_(b)^(@)`
`0.052=T_(b)-100,T_(b)=100.052`