A \(1.60-\mathrm{g}\) sample of a mixture of naphthalene \(\left(\mathrm{C}_{10} \mathrm{H}_{8}\right)\) and anthracene \(\left(\mathrm{C}_{14} \mathrm{H}_{10}\right)\) is dissolved in \(20.0 \mathrm{g}\) benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right) .\) The freezing point of the solution is \(2.81^{\circ} \mathrm{C} .\) What is the composition as mass percent of the sample mixture? The freezing point of benzene is \(5.51^{\circ} \mathrm{C}\) and \(K_{\mathrm{f}}\) is \(5.12^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol}\)

First, we will find the freezing point depression \(\Delta T_f\) by subtracting the freezing point of the solution from the freezing point of pure benzene: \(\Delta T_f = T_{f, pure} - T_{f, solution} = 5.51 - 2.81 = 2.7^{\circ}\text{C}\)

We will use the freezing point depression equation to determine the molality of the solution: \(\Delta T_f = K_f \cdot m\) From the given data, we know that \(K_f = 5.12^{\circ}\text{C}\cdot\text{kg/mol}\) and \(\Delta T_f = 2.7^{\circ}\text{C}\). Therefore, the molality of the solution (m) can be calculated as follows: \(m = \frac{\Delta T_f}{K_f} = \frac{2.7}{5.12} = 0.5273\, \text{mol/kg}\)

Now, we can calculate the moles of solute (mixture of naphthalene and anthracene) using the molality and the mass of the solvent (benzene). Moles of solute = Molality x Mass of solvent Moles of solute = \(0.5273 \, \text{mol/kg} \cdot 0.020\, \text{kg} = 0.010546 \, \text{mol}\)

Let \(x_1\) be the mole fraction of naphthalene and \(x_2\) be the mole fraction of anthracene. The sum of the two mole fractions is equal to 1. Since the total mass of the mixture is 1.60 g, we can write two equations: \(x_1 M_1 + x_2 M_2 = 0.010546 \, \text{mol}\) - Equation 1 (Mole sum equation) \(x_1 m_1 + x_2 m_2 = 1.60 \, \text{g}\) - Equation 2 (Mass sum equation) Here, \(M_1\) and \(M_2\) are the molar masses of naphthalene and anthracene, respectively, and \(m_1\) and \(m_2\) are the masses of naphthalene and anthracene, respectively.

Solving Equation 1 for \(x_1\), we get: \(x_1 = \frac{0.010546 \, \text{mol} - x_2 M_2}{M_1}\) Substituting this into Equation 2, we get: \(\left(\frac{0.010546 \, \text{mol} - x_2 M_2}{M_1}\right) m_1 + x_2 m_2 = 1.60 \, \text{g}\) Next, we substitute the molar masses of naphthalene (\(M_1 = 128.17 \, \text{g/mol}\)) and anthracene (\(M_2 = 178.23 \, \text{g/mol}\)) and solve for \(x_2\): \(\left(\frac{0.010546 - x_2 (178.23)}{128.17}\right) (128.17) + x_2 (178.23) = 1.60\) This equation can be solved numerically to find the mole fraction of anthracene (\(x_2\)): \(x_2 = 0.2557\) Now, we can find the mole fraction of naphthalene (\(x_1\)) by subtracting the mole fraction of anthracene from 1: \(x_1 = 1 - x_2 = 1 - 0.2557 = 0.7443\)

To determine the mass percent for each component, we will first calculate the masses of naphthalene and anthracene using the mole fractions and the molar masses: Mass of naphthalene (\(m_1\)) = \(x_1 \cdot M_1\) = \(0.7443 \cdot 128.17\) = 95.37 g Mass of anthracene (\(m_2\)) = \(x_2 \cdot M_2\) = \(0.2557 \cdot 178.23\) = 45.54 g Finally, we will find the mass percent composition: Mass percent of naphthalene = \(\frac{95.37}{95.37 + 45.54} \times 100 = 67.66\%\) Mass percent of anthracene = \(\frac{45.54}{95.37 + 45.54} \times 100 = 32.34\%\) The composition of the mixture is approximately \(67.66\%\) naphthalene and \(32.34\%\) anthracene by mass.