Answer
See the answer below.
Work Step by Step
a) Freezing point depression: $-16°C$
Molality: $\Delta T_f=K_fm\rightarrow -16=-1.86m\rightarrow m=8.60\ m$
b) For 1 kg of water: $1\ kg\times 8.60\ m=8.60\ mol$ of ethanol
Mass of solute: $8.60\ mol\times 46.07\ g/mol=396.30\ g$
Weight percent: $396.6/(1000+396.3)\times100\%=0.29$
