Question

How many grams of methyl alcohol should be added to $10$ litre tank of water to prevent its freezing at $268\, K$ ? $\left(K_{f} \, for\, water\, is \,1.86 K\, kg \, mol^{-1} \right)$

• A. 880.07 g
• B. 899.04 g
• C. 886.02 g
• D. 868.06 g

Answer 1

The Correct Option is D

To determine the amount of methyl alcohol needed to prevent freezing of water at a given temperature, we use the concept of freezing point depression. The formula for freezing point depression is given by:

\Delta T_f = K_f \cdot \text{molality}

Here, \Delta T_f is the decrease in freezing point, K_f is the cryoscopic constant for water, and molality is the number of moles of solute per kilogram of solvent.

1. Initial freezing point of pure water is 273 \, K. The desired freezing point is 268 \, K.
2. Therefore, the decrease in freezing point, \Delta T_f = 273 \, K - 268 \, K = 5 \, K.
3. Using the formula: 5 = 1.86 \cdot \text{molality}, we solve for molality:

\text{molality} = \frac{5}{1.86} \approx 2.688 \, \text{mol/kg}

4. Methyl alcohol (CH₃OH) has a molar mass of about 32 \, \text{g/mol}.
5. The molality formula gives us moles of solute per kg of solvent. Here, the solvent is water, and its mass is 10 \, \text{kg} because the density of water is approximately 1 \, \text{kg/L}.

The number of moles of methyl alcohol required:

\text{Moles} = 2.688 \, \text{mol/kg} \cdot 10 \, \text{kg} = 26.88 \, \text{mol}

6. Convert moles to grams:
   26.88 \, \text{mol} \times 32 \, \text{g/mol} = 860.16 \, \text{g}

Hence, approximately 868.06 g of methyl alcohol should be added to 10 L of water.