Answer
$-10.55\ ^{\circ}C$
Work Step by Step
Molality: $(35.0\ g\div 110.98\ g/mol)\div 0.150\ kg=2.10\ mol/kg$
F.p. depression: $\Delta T_f=i.K_f.m=2.7\times-1.86^{\circ}C/m\times 2.10\ m=-10.55\ ^{\circ}C$
Freezing point: $-10.55\ ^{\circ}C$