Question

What is the boiling point and freezing point of a 2.21m AlCl3 aqueous solution? (the kf=1.86C/m and kb=0.512C/m)

Answer 1

ΔTb =  i  * Kb * m

ΔTb = 4 * 0.512 C/m * 2.21 m

solve DTb and add it to the BP of pure water

i get a new BP of 104.5 C and new FP of - 16.4 C

hope this helps