Answer
See the answer below.
Work Step by Step
a) Osmotic pressure: $\Pi=cRT\rightarrow (3.8/760)atm=c\times0.082\ L.atm/mol.K\times (273+25)K$
$c=2.046\times10^{-4}\ mol/L$
Molar mass: $10.0\ g/L\div 2.046\times10^{-4}\ mol/L=48872\ g/mol$
b) Assuming $1\ L \approx 1\ kg_{solvent}\rightarrow m=2.046\times10^{-4}\ mol/kg$
F.p. depression: $\Delta T_f=K_fm=-1.86^{\circ}C/m\times 2.046\times10^{-4}\ mol/kg= -0.00038\ ^{\circ}C$
Freezing point: $-0.00038\ ^{\circ}C$
Since the value is very small, you would either need very precise equipment to measure it or incur severe experimental error.
