A solution is prepared by dissolving12.1g of sucrose (C12H22O11, 342.3 g/mol) in170.g of water. Calculate the depression of the freezing point.
(Kf for water is1.86°C kg/mol). Sucrose is a nonelectrolyte.

To calculate the depression of the freezing point of the solution, we use the formula:

\[
\Delta T_f = K_f \times m
\]

where:
- \(\Delta T_f\) is the freezing point depression,
- \(K_f\) is the cryoscopic constant (freezing point depression constant) for the solvent (water, in this case, which is \(1.86 \, \text{°C kg/mol}\)),
- \(m\) is the molality of the solution.

### Step 1: Calculate the molality of the solution

Molality (\(m\)) is defined as the number of moles of solute per kilogram of solvent:

\[
m = \frac{\text{moles of solute}}{\text{kilograms of solvent}}
\]

**Determine the moles of sucrose (solute):**

\[
\text{Moles of sucrose} = \frac{\text{mass of sucrose}}{\text{molar mass of sucrose}}
\]

Given:
- Mass of sucrose = \(12.1 \, \text{g}\)
- Molar mass of sucrose = \(342.3 \, \text{g/mol}\)

\[
\text{Moles of sucrose} = \frac{12.1}{342.3} \approx 0.0354 \, \text{mol}
\]

**Convert the mass of the solvent (water) to kilograms:**

\[
\text{Mass of water} = 170.0 \, \text{g} = 0.170 \, \text{kg}
\]

**Calculate the molality:**

\[
m = \frac{0.0354 \, \text{mol}}{0.170 \, \text{kg}} \approx 0.208 \, \text{mol/kg}
\]

### Step 2: Calculate the freezing point depression

Now, substitute the values into the formula for freezing point depression:

\[
\Delta T_f = K_f \times m = 1.86 \, \text{°C kg/mol} \times 0.208 \, \text{mol/kg}
\]

\[
\Delta T_f \approx 0.387 \, \text{°C}
\]

### Final Answer

The depression of the freezing point is approximately **0.387°C**.

Would you like a detailed explanation of each step, or do you have any other questions?

Here are 5 related questions to consider:
1. What would happen to the freezing point depression if the mass of sucrose was doubled?
2. How does the freezing point depression change with different solutes?
3. Can you explain why sucrose is considered a nonelectrolyte?
4. What is the impact of using a different solvent with a higher or lower \(K_f\) value?
5. How does molality differ from molarity, and when is each used?

**Tip:** Remember that nonelectrolytes like sucrose do not dissociate into ions in solution, which keeps their effect on freezing point depression straightforward compared to electrolytes.

Learn how to calculate the freezing point depression of a sucrose solution using the formula \( \Delta T_f = K_f \times m \). Detailed step-by-step solution provided with specific data points. Suitable for high school students studying colligative properties.

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