Answer
$T_f=-29.9C^{\circ}, T_b=108.2C^{\circ}$
Work Step by Step
We can find the molality as
$m=\frac{50.0g C_2H_6O_2\times \frac{1mol}{62.07g}}{50.0gH_2O\times \frac{1Kg}{1000g}}=16.1\frac{mol}{Kg}$
Now $\Delta T_f=K_fm$
$\Delta T_f=1.86\times 16.1=29.9C^{\circ}$
$T_f=0.0-29.9=-29.9C^{\circ}$
$\Delta T_b=K_bm$
We plug in the known values to obtain:
$\Delta T_b=0.51\times 16.1=8.2C^{\circ}$
$T_b=100.0+8.2=108.2C^{\circ}$