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Ch.13 - Solutions & Their Properties
All textbooksMcMurry 8th EditionCh.13 - Solutions & Their PropertiesProblem 143
Chapter 13, Problem 143

A solution of 0.250 g of naphthalene (mothballs) in 35.00 g of camphor lowers the freezing point by 2.10 °C. What is the molar mass of naphthalene? The freezing-point-depression constant for camphor is 37.7 °C kg/mol.

Verified step by step guidance
1
Identify the formula for freezing point depression: \( \Delta T_f = i \cdot K_f \cdot m \), where \( \Delta T_f \) is the change in freezing point, \( i \) is the van't Hoff factor (which is 1 for naphthalene as it does not ionize), \( K_f \) is the freezing-point-depression constant, and \( m \) is the molality of the solution.
Rearrange the formula to solve for molality \( m \): \( m = \frac{\Delta T_f}{K_f} \). Substitute the given values: \( \Delta T_f = 2.10 \text{ °C} \) and \( K_f = 37.7 \text{ °C kg/mol} \).
Calculate the molality \( m \) using the rearranged formula.
Use the definition of molality: \( m = \frac{\text{moles of solute}}{\text{kg of solvent}} \). Here, the mass of camphor is 35.00 g, which is 0.035 kg.
Solve for the moles of naphthalene using the calculated molality and the mass of camphor. Then, use the relationship \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \) to find the molar mass of naphthalene, given the mass of naphthalene is 0.250 g.
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