$$\Delta T_f = K_f * m$$
where \(\Delta T_f\) is the freezing point depression, \(K_f\) is the freezing point depression constant of the solvent (\(K_f = 1.86 °C/m\) for water), and \(m\) is the molality of the solution.
Rearranging to solve for \(m\):
$$m = \frac{\Delta T_f}{K_f}$$
First, we need to calculate the freezing point depression:
$$\Delta T_f = -10.0 °C - 0.0 °C (Initial temperature of water is 0°C) = -10.0 °C $$
Now we can calculate the molality:
$$m = \frac{-10.0 °C}{1.86 °C/m} = -5.38 m$$
To find the grams of NaCl needed, we need to use the formula relating molality to the number of moles and mass of solute:
$$m = \frac{moles\ of\ NaCl}{kg\ of\ solvent}$$
Rearranging to solve for the moles of NaCl:
$$moles \ of \ NaCl = m * kg\ of\ solvent$$
Converting grams to kilograms:
$$moles \ of \ NaCl= (-5.38\ m) * 3.5 kg = -18.83\ moles \ of \ NaCl $$
Finally converting moles to grams:
$$-18.83\ moles \ of \ NaCl * (58.44 g/mol) = \boxed{-1100\ g \ NaCl }$$
(since the molecular mass of NaCl is 58.44 g/mol)
Therefore, -1100 g of NaCl must be added to 3.5 kg (3500) grams of water to reach a temperature of -10.0°C.
