Answer
(a) $100.060^{\circ}\,C$
(b) $100.993^{\circ}\,C$
(c) $101.99^{\circ}C$
(d) $101.11^{\circ}C$
Work Step by Step
(a) $\Delta T_{b}=m\times K_{b}$
Molality $m=0.118\,m=0.118\,\frac{mol\,solute}{kg\,solvent}$
Boiling point elevation constant for water is $K_{b}=0.512\,\frac{^{\circ}C\,kg\,solvent}{mol\,solute}$
Then $\Delta T_{b}=0.118\,\frac{mol\,solute}{kg\,solvent}\times0.512\,\frac{^{\circ}C\,kg\,solvent}{mol\,solute}$
$=0.0604^{\circ}C$
Boiling point= $100.000^{\circ}C+0.0604^{\circ}C=100.060^{\circ}C$
(b) $m=1.94\,\frac{mol\,solute}{kg\,solvent}$
$\Delta T_{b}=1.94\,\frac{mol\,solute}{kg\,solvent}\times0.512\,\frac{^{\circ}C\,kg\,solvent}{mol\,solute}$
$=0.993^{\circ}C$
Boiling point= $100.000^{\circ}C+0.993^{\circ}C=100.993^{\circ}C$
(c) $m=3.88\,\frac{mol\,solute}{kg\,solvent}$
$\Delta T_{b}=3.88\,\frac{mol\,solute}{kg\,solvent}\times0.512\,\frac{^{\circ}C\,kg\,solvent}{mol\,solute}$
$=1.99^{\circ}C$
Boiling point= $100.000^{\circ}C+1.99^{\circ}C=101.99^{\circ}C$
(d) $m=2.16\,\frac{mol\,solute}{kg\,solvent}$
$\Delta T_{b}=2.16\,\frac{mol\,solute}{kg\,solvent}\times0.512\,\frac{^{\circ}C\,kg\,solvent}{mol\,solute}$
$=1.11^{\circ}C$
Boiling point= $100.000^{\circ}C+1.11^{\circ}C=101.11^{\circ}C$