Answer
See the answer below.
Work Step by Step
In 1 kg of the solution:
Number of moles of solute: $(2/100\times 1000)g\div256.77\ g/mol=0.078\ mol$
Molality: $m=0.078\ mol\div 0.98\ kg=0.079\ m$
Freezing point depression:
$\Delta T_f=iK_fm\rightarrow i=-0.237^{\circ}C\div(-1.86^{\circ}C/m\times 0.079\ m)=1.60$
There are 1.6 moles of ions in solution per mole of compound.
