Question

What should be the freezing point of aqueous solution containing 17 gm of C₂H₅OH is 100 gm of water. (Kf for water = 1.86)

Answer 1

-6.87 °C

Answer 2

The freezing point of an aqueous solution is given by the formula:

$T_f = T_f^0 - \Delta T_f$

where $T_f^0$ is the freezing point of the pure solvent (water), and $\Delta T_f$ is the depression in freezing point. The depression in freezing point is a colligative property given by:

$\Delta T_f = K_f \times m$

where $K_f$ is the cryoscopic constant of the solvent, and $m$ is the molality of the solution.

First, calculate the molality ($m$) of the solution. Molality is defined as the number of moles of solute per kilogram of solvent. The solute is C₂H₅OH (ethanol). Mass of solute = 17 gm. The molar mass of C₂H₅OH is calculated from the atomic masses: Molar mass of C₂H₅OH = (2 × atomic mass of C) + (6 × atomic mass of H) + (1 × atomic mass of O) Molar mass of C₂H₅OH = (2 × 12.01 g/mol) + (6 × 1.008 g/mol) + (1 × 15.999 g/mol) Molar mass of C₂H₅OH = 24.02 g/mol + 6.048 g/mol + 15.999 g/mol = 46.067 g/mol. Using the approximate value 46 g/mol: Number of moles of C₂H₅OH = $\frac{\text{Mass of solute}}{\text{Molar mass of solute}} = \frac{17 \text{ gm}}{46 \text{ gm/mol}}$.

The solvent is water. Mass of solvent = 100 gm. Convert the mass of solvent to kilograms: Mass of solvent in kg = $\frac{100 \text{ gm}}{1000 \text{ gm/kg}} = 0.1 \text{ kg}$.

Now, calculate the molality ($m$): $m = \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}} = \frac{17/46 \text{ mol}}{0.1 \text{ kg}} = \frac{17}{46 \times 0.1} \text{ mol/kg} = \frac{17}{4.6} \text{ mol/kg}$. $m = \frac{170}{46} \text{ mol/kg} = \frac{85}{23} \text{ mol/kg} \approx 3.69565 \text{ mol/kg}$.

The cryoscopic constant for water ($K_f$) is given as 1.86 °C kg/mol. Calculate the depression in freezing point ($\Delta T_f$): $\Delta T_f = K_f \times m = 1.86 \text{ °C kg/mol} \times \frac{170}{46} \text{ mol/kg}$. $\Delta T_f = 1.86 \times \frac{170}{46} = \frac{316.2}{46} \approx 6.8739 \text{ °C}$.

The freezing point of pure water ($T_f^0$) is 0 °C. The freezing point of the solution ($T_f$) is: $T_f = T_f^0 - \Delta T_f = 0 \text{ °C} - 6.8739 \text{ °C} = -6.8739 \text{ °C}$.

Rounding to two decimal places, the freezing point is -6.87 °C.