Answer
$100.256^{\circ}C$
Work Step by Step
Molality: $(15.0\ g\div60.06\ g/mol)\div 0.500\ kg=0.50\ m$
Boiling Point Elevation: $\Delta T_b=K_b\times m=0.5121^{\circ}C/m\times 0.5\ m=0.256\ ^{\circ}C$
Mixture boiling point: $100.00+0.256=100.256^{\circ}C$
