Answer
1.9
Work Step by Step
We find:
Molality $M=\frac{\text{Moles solute}}{\text{Kilograms solvent}}$
$=\frac{\frac{9.12\,g}{36.46\,g/mol}}{0.190\,kg}=1.3165\,mol/kg$
$\Delta T_{f}=- K_{f}\cdot m \cdot i $
$\implies i=-\frac{\Delta T_{f}}{K_{f}\cdot m}$
$=-\frac{-4.65^{\circ}C}{(1.86^{\circ}C\,kg/mol)(1.3165\,mol/kg)}$
$=1.9$
