Answer
$106.8C^{\circ}$
Work Step by Step
We know that
$m=\frac{\Delta T_f}{m}$
$\implies m=\frac{25}{1.86}=13.4mol\space C_2H_6O_2Kg$
Now $15.0L\space water \times \frac{1.00Kg}{1L}\times \frac{134mol\space C_2H_6O_2}{1KgH_2O}=201mols$
We can find the volume of $C_2H_6O_2$ as follows
$V=201mol \times \frac{62.1}{1mol\space C_2H_6O_2}\times \frac{1mL}{1.11g}=11200mL=11.2L$
$\Delta T_b=K_bm$
We plug in the known values to obtain:
$\Delta T_b=0.51(13.4)=6.8C^{\circ}$
$T_b=100+6.8=106.8C^{\circ}$
