Lauryl alcohol is obtained from coconut oil and found in many hair shampoos. A solution of 5.00 g of lauryl alcohol in 100.0 g of benzene boils at 80.78 ° C. What is the molar mass of lauryl alcohol if k_b = 2.53 ° C/m and the normal boiling point of benzene is 80.10 ° C?

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Quizplus · Accepted Answer

The boiling point elevation formula is $\Delta T_b = i \cdot K_b \cdot m$, where $\Delta T_b$ is the boiling point elevation, $i$ is the van't Hoff factor (which is 1 for non-electrolytes like lauryl alcohol), $K_b$ is the ebullioscopic constant of the solvent (2.53 °C/m for benzene), and $m$ is the molality of the solution. The molality $m$ is defined as moles of solute per kilogram of solvent. The boiling point elevation $\Delta T_b$ is the difference between the boiling point of the solution and the pure solvent, which is $80.78 °C - 80.10 °C = 0.68 °C$.Solving for the molar mass of lauryl alcohol, we first find the molality $m = \frac{\Delta T_b}{K_b} = \frac{0.68}{2.53} = 0.269 mol/kg$. Since we have 5.00 g of lauryl alcohol in 0.100 kg of benzene, we can calculate the molar mass $M$ of lauryl alcohol as $M = \frac{mass\ of\ solute}{moles\ of\ solute} = \frac{5.00\ g}{0.269 \cdot 0.100\ kg} = \frac{5.00}{0.0269} = 186 g/mol$, which is closest to the given option C) 1.9×10^2 g/mol.

Lauryl Alcohol Is Obtained from Coconut Oil and Found in Many