Lauryl alcohol is obtained from coconut oil and is used to make detergents. A solution of 5.00 g of lauryl alcohol in 0.100 kg of benzene freezes at \(4.1^{\circ} \mathrm{C} .\) What is the molar mass of lauryl alcohol from this data?

We are given the following information: - Mass of lauryl alcohol: 5.00 g - Mass of benzene: 0.100 kg - Freezing point depression: 4.1°C

First, we need to calculate the molality (m) of the solution. Molality is defined as the number of moles of solute per kg of solvent. The formula to calculate molality is: \(m = \frac{moles\ of\ solute}{kg\ of\ solvent}\) By re-arranging this formula, we will get: \(Molar\ mass\ of\ lauryl\ alcohol = \frac{Mass\ of\ lauryl\ alcohol}{(molality)(kg\ of\ solvent)}\) We will now use the formula to calculate the molality of the solution: \(\Delta T_f = K_f \cdot m\) Where: - ΔTf is the freezing point depression (4.1°C) - Kf (benzene) is the cryoscopic constant of benzene (5.12 °C kg/mol) - m is the molality of the solution Rearranging the formula, we get: \(m = \frac{\Delta T_f}{K_f}\) Now, plug in the values: \(m = \frac{4.1\,^{\circ}\mathrm{C}}{5.12\,^{\circ}\mathrm{C\,kg/mol}} = 0.800\,\mathrm{mol/kg}\)

Now, we will use the rearranged formula from Step 2 to calculate the molar mass of lauryl alcohol: \(Molar\ mass\ of\ lauryl\ alcohol = \frac{Mass\ of\ lauryl\ alcohol}{(molality)(kg\ of\ solvent)}\) Plugging in the values, we get: \(Molar\ mass\ of\ lauryl\ alcohol = \frac{5.00\,g}{(0.800\,\mathrm{mol/kg})(0.100\,kg)} = \frac{5.00\,g}{0.0800\,mol} = 62.5\,g/mol\) So, the molar mass of lauryl alcohol is approximately 62.5 g/mol.