Answer
(a) $100.115^{\circ}C$
(b) $101.32^{\circ}C$
(c) $102.22^{\circ}C$
(d) $103.47^{\circ}C$
Work Step by Step
(a) $\Delta T_{b}=m\times K_{b}$
Molality $m=0.225\,m=0.225\,\frac{mol\,solute}{kg\,solvent}$
Boiling point elevation constant for water is $K_{b}=0.512\,\frac{^{\circ}C\,kg\,solvent}{mol\,solute}$
Then $\Delta T_{b}=0.225\,\frac{mol\,solute}{kg\,solvent}\times0.512\,\frac{^{\circ}C\,kg\,solvent}{mol\,solute}$
$=0.115^{\circ}C$
Boiling point= $100.000^{\circ}C+0.115^{\circ}C=100.115^{\circ}C$
(b) $m=2.58\,\frac{mol\,solute}{kg\,solvent}$
$\Delta T_{b}=2.58\,\frac{mol\,solute}{kg\,solvent}\times0.512\,\frac{^{\circ}C\,kg\,solvent}{mol\,solute}$
$=1.32^{\circ}C$
Boiling point= $100.000^{\circ}C+1.32^{\circ}C=101.32^{\circ}C$
(c) $m=4.33\,\frac{mol\,solute}{kg\,solvent}$
$\Delta T_{b}=4.33\,\frac{mol\,solute}{kg\,solvent}\times0.512\,\frac{^{\circ}C\,kg\,solvent}{mol\,solute}$
$=2.22^{\circ}C$
Boiling point= $100.000^{\circ}C+2.22^{\circ}C=102.22^{\circ}C$
(d) $m=6.77\,\frac{mol\,solute}{kg\,solvent}$
$\Delta T_{b}=6.77\,\frac{mol\,solute}{kg\,solvent}\times0.512\,\frac{^{\circ}C\,kg\,solvent}{mol\,solute}$
$=3.47^{\circ}C$
Boiling point= $100.000^{\circ}C+3.47^{\circ}C=103.47^{\circ}C$
