QJEEM24Ch84b5: JEE Main 2024 Chemistry - Class 12 JEEM24Ch84b5 Solution: Mass of ethylene glycol (antifreeze) to be added to of water to …

To determine the mass of ethylene glycol (antifreeze) needed to lower the freezing point of water to $-24^{\circ} \mathrm{C}$ , we can use the freezing point depression equation, which is a colligative property given by:

where:

• $\Delta T_f$ is the depression in the freezing point.
• $i$ is the van't Hoff factor, which is the number of particles the solute splits into in solution. For ethylene glycol, which does not dissociate in solution, $i = 1$ .
• $K_f$ is the cryoscopic constant (freezing point depression constant) of water, which is given as $1.86 \mathrm{~K} ~\mathrm{kg} ~\mathrm{mol}^{-1}$ .
• $m$ is the molality of the solution.

However, molality is defined as the number of moles of solute per kilogram of solvent, so we first need to find the molality that corresponds to the desired freezing point depression:

Now, we can solve for $m$ :

Substituting the given values:

Note that $\mathrm{C}$ and $\mathrm{K}$ are interchangeable when calculating changes in temperature. Now, we can compute the molality:

Next, we calculate the molality of the ethylene glycol:

Keep in mind that molality is always positive, but the negative sign indicates the direction of the temperature change. Since we're calculating the amount needed, we can consider it as 12.903 moles per kilogram of water.

Now we calculate the moles of ethylene glycol required for $18.6 \mathrm{~kg}$ of water:

Next, we find the mass of ethylene glycol needed using its molar mass:

Given the molar mass of ethylene glycol is $62 \mathrm{~g} ~\mathrm{mol}^{-1}$ , we then convert it to $\mathrm{kg} ~\mathrm{mol}^{-1}$ by dividing by 1000:

Now we can calculate the mass:

Therefore, approximately $14.881 \mathrm{~kg}$ of ethylene glycol needs to be added to $18.6 \mathrm{~kg}$ of water to lower the freezing point to $-24^{\circ} \mathrm{C}$ .