Question

A solution of 'P' moles of sucrose in 100 g of water freezes at -0.2°C. As ice separates out, the freezing point goes down to -0.25°C. How many grams of ice would have separated?

Answer 1

20 g

Answer 2

• Concept: The depression in freezing point ($\Delta T_f$) is a colligative property that depends on the molality ($m$) of the solution. It is given by the formula:

  $\Delta T_f = K_f \times m$

  where $K_f$ is the cryoscopic constant for the solvent (for water, $K_f = 1.86 \text{ K kg mol}^{-1}$). Molality ($m$) is defined as moles of solute per kilogram of solvent.

• Initial State: The initial freezing point of the solution is $-0.2^\circ C$. The freezing point of pure water is $0^\circ C$. So, the depression in freezing point, $\Delta T_{f1} = 0^\circ C - (-0.2^\circ C) = 0.2^\circ C$. The mass of water (solvent) is 100 g = 0.1 kg. Let 'P' be the moles of sucrose. The molality of the initial solution, $m_1 = \frac{P \text{ mol}}{0.1 \text{ kg}}$. Using the formula:

  $0.2 = 1.86 \times \frac{P}{0.1}$

  $P = \frac{0.2 \times 0.1}{1.86} = \frac{0.02}{1.86} \text{ mol}$

• Final State (after ice separates): The freezing point of the solution goes down to $-0.25^\circ C$. The new depression in freezing point, $\Delta T_{f2} = 0^\circ C - (-0.25^\circ C) = 0.25^\circ C$. The moles of sucrose 'P' remain constant in the solution. Let $w'$ be the mass of water remaining in the solution in grams. So, the mass of water in kg is $\frac{w'}{1000}$ kg. The molality of the final solution, $m_2 = \frac{P \text{ mol}}{w'/1000 \text{ kg}}$. Using the formula:

  $0.25 = 1.86 \times \frac{P}{w'/1000}$

  Substitute the value of P from the initial state:

  $0.25 = 1.86 \times \frac{0.02/1.86}{w'/1000}$

  $0.25 = \frac{0.02}{w'/1000}$

  $0.25 = \frac{0.02 \times 1000}{w'}$

  $0.25 = \frac{20}{w'}$

  $w' = \frac{20}{0.25} = 80 \text{ g}$ This is the mass of water remaining in the solution.

• Mass of Ice Separated: Initial mass of water = 100 g Mass of water remaining = 80 g Mass of ice separated = Initial mass of water - Mass of water remaining Mass of ice separated = $100 \text{ g} - 80 \text{ g} = 20 \text{ g}$

The final answer is $\boxed{\text{20g}}$.

Explanation of the solution:

1. Calculate initial freezing point depression ($\Delta T_{f1}$) and use it with the cryoscopic constant ($K_f$) and initial water mass to find the moles of sucrose (P).
2. Calculate the new freezing point depression ($\Delta T_{f2}$) after ice separation.
3. Use $\Delta T_{f2}$, $K_f$, and the calculated moles of sucrose (P) to determine the mass of water remaining in the solution.
4. Subtract the mass of remaining water from the initial mass of water to find the mass of ice separated.