Use freezing point depression formula

$\Delta T_f = iK_f m$

For ethylene glycol, there is no dissociation, so

$i=1$

Given:

• $K_f = 1.86\,\text{K kg mol}^{-1}$
• Lowering in freezing point $= 2.8^\circ\text{C}$, so $\Delta T_f = 2.8\,\text{K}$

Calculate molality

$m = \frac{\Delta T_f}{K_f} = \frac{2.8}{1.86}$

$m \approx 1.505\,\text{mol kg}^{-1}$

Find moles of ethylene glycol needed

Molality is moles of solute per kg of solvent.

Since radiator contains $1.0\,\text{kg}$ of water,

$\text{moles of ethylene glycol} = 1.505 \times 1.0 = 1.505\,\text{mol}$

Calculate molar mass of ethylene glycol $C_2H_6O_2$

$M = 2(12) + 6(1) + 2(16) = 24 + 6 + 32 = 62\,\text{g mol}^{-1}$

Calculate required mass

$\text{mass} = \text{moles} \times \text{molar mass}$

$\text{mass} = 1.505 \times 62$

$\text{mass} \approx 93.3\,\text{g}$

Match with options

Closest option is:

$\boxed{93\,\text{g}}$

So the correct option is B.

Comparison with stored answer

Stored correct answer: B

Derived answer: B

Hence, the answers agree.