Answer
$-24C^{\circ}$
Work Step by Step
We can find the required freezing point as follows:
$\Delta T_b=106.5C^{\circ}-100.0C^{\circ}=6.5C^{\circ}$
$m=\frac{\Delta T_b}{K_b}$
$\implies m=\frac{6.5C^{\circ}}{0.512\frac{C^{\circ}}{m}}=12.695m$
Now $\Delta T_f=K_f\times m$
$\Delta T_f=(1.86\frac{C^{\circ}}{m})(12.695m)=24C^{\circ}$
$freezing \space point=0C^{\circ}-\Delta T_f$
$freezing \space point=0C^{\circ}-24C^{\circ}=-24C^{\circ}$
