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Chapter 11: Problem 67

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

Short Answer

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

Step by step solution

01

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)
02

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): \(m = \frac{\Delta T_{f}}{K_{f}}\)
03

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掳C is 0.0252 mol/kg.

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