$K_{\mathrm{f}}$ r for water is $1.86\mathrm{Kg} {\mathrm{mol}}^{-1}$. If your automobile radiator holds 1.0 kg of water, how many grams of ethylene glycol $\left({\mathrm{C}}_2{\mathrm{H}}_6{\mathrm{O}}_2\right)$ must you add to get the freezing point of hte solution lowered to - 2.8 °C?

Since $-\mathrm{\Delta }{T}_{\mathrm{f}}=K_{\mathrm{f}}m$ we get 

$m=\frac{-\mathrm{\Delta }{T}_{\mathrm{f}}}{K_{\mathrm{f}}}=\frac{2.8\mathrm{K}}{1.86\mathrm{Kkg}{\mathrm{mol}}^{-1}}=1.505\mathrm{mol}$

Since the molality, $m=n/m_1$, we get 

$n=m{m}_1=\left(1.505\mathrm{mol}{\mathrm{kg}}^{-1}\right)\left(1\mathrm{kg}\right)=1.505\mathrm{mol}$

Finally, the mass of ethylene glycol, ${\mathrm{C}}_2{\mathrm{H}}_6{\mathrm{O}}_2$ (the molar mass = 62 g mor ^-1) required will be

$m=n{M}_{\mathrm{m}}=\left(1.505\mathrm{mol}\right)\left(62\mathrm{g}{\mathrm{mol}}^{-1}\right)=93.3\mathrm{g}$