Answer
Freezing point= $-7.44^{\circ}C$
Boiling point= $102.05^{\circ}C$
Work Step by Step
Molality=$ \frac{\text{moles solute}}{\text{kilograms solvent}}$
$=\frac{\frac{21.2\,g}{62.07\,g/mol}}{1.00\times10^{-3}\,kg/mL\times85.4\,mL }=4.00\,m $
$\Delta T_{f}=m\times K_{f}=(4.00\times1.86)^{\circ}C=7.44^{\circ}C$
Freezing point= $0.00^{\circ}C-7.44^{\circ}C$
$=-7.44^{\circ}C$
$\Delta T_{b}=m\times K_{b}=(4.00\times0.512)^{\circ}C$
$=2.05^{\circ}C$
Boiling point= $100.00^{\circ}C+2.05^{\circ}C$
$=102.05^{\circ}C$
