Answer
$-3.55^{\circ}C$
Work Step by Step
We find:
Molality $M=\frac{\text{Moles solute}}{\text{Kilograms solvent}}$
$=\frac{\frac{7.40\,g}{95.211\,g/mol}}{0.110\,kg}=0.70656\,mol/kg$
$\Delta T_{f}=- K_{f}\cdot m \cdot i $
$=-(1.86^{\circ}C\,kg/mol)(0.70656\,mol/kg)(2.7)$
$=-3.548^{\circ}C$
Freezing point= $0.00^{\circ}C+(-3.548^{\circ}C)=-3.55^{\circ}C$
