Answer
See the answer below.
Work Step by Step
In 1 kg of the solution:
Mass of solute:
$3.00\%\times 1000\ g=30\ g$
Number of moles of solute:
$30\ g\div 165.19\ g/mol=0.182\ mol$
Molality:
$0.182\ mol\div 0.970\ kg=0.187\ m$
Assuming density is 1.0 kg/L
$c=0.182\ mol/L$
a) Freezing point depression:
$\Delta T_f=K_fm=-1.86^{\circ}C/m\times 0.187\ m=-0.35^{\circ}C$
Freezing point $0-0.35=-0.35^{\circ}C$
b) Boiling point elevation:
$\Delta T_b=K_bm=0.5121^{\circ}C/m\times 0.187\ m=0.096^{\circ}C$
Boiling point
$100+0.096=100.096^{\circ}C$
c) Osmotic pressure
$\Pi=cRT=0.182\ mol/L\times 0.082\ L.atm/mol.K\times (25+273)K=4.44\ atm$
Since the osmotic pressure is very large, it's the one that can be measured the easiest with a small error.
