Equal volumes of ethylene glycol (molar mass = 62) and water (molar mass = 18) are mixed. The depression in the freezing point of water is (given K_f of water = 1.86 K mol^–1 kg and the specific gravity of ethylene glycol is 1.11)

33.3

The correct answer is option 4. 33.3.

To calculate the depression in freezing point caused by the addition of ethylene glycol to water, we can use the formula:

\(\Delta T_f = K_f \cdot m \cdot i\)

where:
\(\Delta T_f\) is the depression in freezing point,
\(K_f\) is the cryoscopic constant of the solvent (water),
\(m\) is the molality of the solute (ethylene glycol), and
\(i\) is the Van't Hoff factor, which represents the number of particles formed when the solute dissolves.

First, we need to calculate the molality of ethylene glycol. Molality (\(m\)) is defined as the moles of solute per kilogram of solvent. Since equal volumes are mixed, we can assume we have 1 kg of water.

The molar mass of ethylene glycol is 62 g/mol, and the specific gravity of ethylene glycol is 1.11. Therefore, the mass of ethylene glycol required to have the same volume as water is:

\(\text{mass of ethylene glycol} = \text{specific gravity of ethylene glycol} \times \text{mass of water}\)
\(\text{mass of ethylene glycol} = 1.11 \times 1000 \, \text{g} = 1110 \, \text{g}\)

Converting the mass of ethylene glycol to moles:

\(\text{moles of ethylene glycol} = \frac{\text{mass of ethylene glycol}}{{\text{molar mass of ethylene glycol}}}\)
\(\text{moles of ethylene glycol} = \frac{1110 \, \text{g}}{{62 \, \text{g/mol}}} \approx 17.903 \, \text{mol}\)

Since we have 1 kg of water, the molality of ethylene glycol is:

\(m = \frac{{\text{moles of solute}}}{{\text{mass of solvent (kg)}}} = \frac{{17.903 \, \text{mol}}}{{1 \, \text{kg}}} = 17.903 \, \text{mol/kg}\)

The Van't Hoff factor (\(i\)) for ethylene glycol is 1 because it does not dissociate or form ions when it dissolves.

Now we can calculate the depression in freezing point:

\(\Delta T_f = K_f \cdot m \cdot i = 1.86 \, \text{K }mol^{-1} \, \text{kg} \cdot 17.903 \, \text{mol/kg} \cdot 1 = 33.331 \, \text{K}\)

The depression in freezing point is 33.331 K, which corresponds to option (4).