## Chemistry: The Molecular Science (5th Edition)

$K_{f}$ for water= $-1.86^{\circ}C\,kg/mol$ Moles of solute n= $\frac{1.20\,kg}{molar\,mass\,of\, ethylene\, glycol}$ $=\frac{1.20\times10^{3}g}{62.068\,g/mol}=19.3\,mol$ Molality $m_{solute}=\frac{moles\,of\,solute}{mass\,of\,solvent\,in\,kg}$ $=\frac{19.3\,mol}{6.50\,kg}=2.97\,mol/kg$ Now, $T_{f}(solution)-T_{f}(solvent)=K_{f}\times m_{solute}$ We know that $T_{f}(solvent)=0^{\circ}C$ as the solvent is water. Then, we have $T_{f}(solution)=( -1.86^{\circ}C)(2.97)=-5.52^{\circ}C$ The solution freezes at $-5.52^{\circ}C$. The addition of 1.20 kg ethylene glycol is not sufficient to prevent freezing if the temperature drops to $-25^{\circ}C$.