#### Answer

Not sufficient. See the explanation below.

#### Work Step by Step

$K_{f}$ 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 $-25^{\circ}C$.