Problem 67 - Chemistry: An Atoms-Focused Approach

Question

What molality of a nonvolatile, nonelectrolyte solute is needed to lower the melting point of camphor by \(1.000^{\circ} \mathrm{C}\) \(\left(K_{f}=39.7^{\circ} \mathrm{C} / m\right) ?\)

Step-by-Step Solution

Answer

Answer: The molality of the nonvolatile, nonelectrolyte solute needed to lower the melting point of camphor by 1.00°C is 0.0252 mol/kg.

Step 1: Identify the given values

We are given the following values: \(\Delta T_{f} = 1.000^{\circ}\mathrm{C}\) (the desired change in freezing point) \(K_{f} = 39.7^{\circ}\mathrm{C}/m\) (the molal freezing-point depression constant)

Step 2: Apply the freezing point depression formula

We will use the freezing point depression formula, \(\Delta T_{f}=K_{f} \cdot m\), to solve for the molality (m). First, rearrange the formula to solve for molality (m): \(m = \frac{\Delta T_{f}}{K_{f}}\)

Step 3: Plug in the given values and solve for molality

Now that we have the formula for molality, insert the given values and calculate: \(m = \frac{1.000^{\circ}\mathrm{C}}{39.7^{\circ}\mathrm{C}/m}\) \(m = 0.0252m\) So, the molality of the nonvolatile, nonelectrolyte solute needed to lower the melting point of camphor by 1.00°C is 0.0252 mol/kg.