83 g of ethylene glycol dissolved in 625 g of water. The freezing point of the solution is _________ K. (Nearest integer)
[Use: Molal Freezing point depression constant of water = 1.86 K kg mol$$^{-1}$$
Freezing point of water = 273 K
Atomic masses: C : 12.0u, O : 16.0u, H : 1.0u]

We start with the relation for freezing-point depression. According to Raoult’s law for dilute solutions, the depression in freezing point $$\Delta T_f$$ is given by the formula

$$\Delta T_f = K_f \, m$$

where $$K_f$$ is the molal freezing-point depression constant of the solvent and $$m$$ is the molality of the solution.

First, we calculate the molality. Molality is defined as

$$m = \dfrac{\text{moles of solute}}{\text{mass of solvent (in kg)}}$$

Our solute is ethylene glycol, whose molecular formula is $$\mathrm{C_2H_6O_2}$$. Using the given atomic masses, we obtain its molar mass.

$$\text{Molar mass of } \mathrm{C_2H_6O_2}=2(\text{C})+6(\text{H})+2(\text{O})$$

$$=2(12.0\ \text{u})+6(1.0\ \text{u})+2(16.0\ \text{u})$$

$$=24\ \text{u}+6\ \text{u}+32\ \text{u}=62\ \text{g mol}^{-1}$$

Now we find moles of ethylene glycol present in 83 g of the compound.

$$n=\dfrac{\text{mass}}{\text{molar mass}}=\dfrac{83\ \text{g}}{62\ \text{g mol}^{-1}}$$

$$n \approx 1.3387\ \text{mol}$$

The mass of water acting as solvent is 625 g, which in kilograms is

$$625\ \text{g}=0.625\ \text{kg}$$

So the molality becomes

$$m=\dfrac{1.3387\ \text{mol}}{0.625\ \text{kg}}$$

$$m \approx 2.1419\ \text{mol kg}^{-1}$$

With the molality in hand, we apply the freezing-point depression formula. The constant given is $$K_f = 1.86\ \text{K kg mol}^{-1}$$.

$$\Delta T_f = K_f \, m = 1.86\ \text{K kg mol}^{-1}\times 2.1419\ \text{mol kg}^{-1}$$

$$\Delta T_f \approx 3.986\ \text{K}$$

The negative sign signifies a lowering of the freezing point, so the new freezing point of water becomes

$$T_f(\text{solution}) = T_f(\text{pure water}) - \Delta T_f$$

$$T_f(\text{solution}) = 273\ \text{K} - 3.986\ \text{K}$$

$$T_f(\text{solution}) \approx 269.014\ \text{K}$$

Rounded to the nearest integer, the freezing point of the solution is 269 K.

So, the answer is $$269\ \text{K}$$.