We know,
Number of moles in 100 ml of 3 M = 0.3 mol
Again, density of pure A = 32 g/ml
So, mass of 100 ml pure A = 32×100=150 g
So molar mass of A (MA)=1500.3=500 g
Again,
number of moles of B in 100 ml, 8 M = 0.8 mol
Given, density of pure B = 2 g/ml
So, mass of 100 ml of pure B = 2×100 g=200 g
So B is the solvent here.
Hence molality of the final solution =0.3200×1000=1.5 m
So our answer will be =1.50.3=5