Answer
See the answer below.
Work Step by Step
The number of moles of ions per gram of water: $\Sigma_i X_i(ppm)\times10^{-6}g/ppm\div M_i(g/mol)=1.103\times10^{-3}\ mol$
Molality: $1.103\times10^{-3}\ mol\div 0.001\ kg=1.103\ m$
Molarity: $\approx1.103\ mol/L$
a) F.p. depression: $\Delta T_f=K_fm=-1.86^{\circ}C/m\times 1.103\ m=-2.05^{\circ}C$
b) Osmotic pressure(25°C): $\Pi=cRT=1.103\ mol/L\times 0.082\ L.atm/mol.K\times (273+25)K=27.0\ atm$
The minimum pressure for reverse osmosis would be 27 atm.
