Kf r for water is 1.86Kg mol−1. If your automobile radiator holds 1.0 kg of water, how many grams of ethylene glycol C2H6O2 must you add to get the freezing point of hte solution lowered to - 2.8 °C?

Since $- Δ T_{f} = K_{f} m$ we get

$m = \frac{- Δ T_{f}}{K_{f}} = \frac{2.8 K}{1.86 Kkg {mol}^{- 1}} = 1.505 mol$

Since the molality, $m = n / m_{1}$ , we get

$n = m m_{1} = (1.505 mol {kg}^{- 1}) (1 kg) = 1.505 mol$

Finally, the mass of ethylene glycol, $C_{2} H_{6} O_{2}$ (the molar mass = 62 g mor ^-1) required will be

$m = n M_{m} = (1.505 mol) (62 g {mol}^{- 1}) = 93.3 g$

$K_{f}$ r for water is $1.86 Kg {mol}^{- 1}$ . If your automobile radiator holds 1.0 kg of water, how many grams of ethylene glycol $(C_{2} H_{6} O_{2})$ must you add to get the freezing point of hte solution lowered to - 2.8 °C?