Question

1. what is the molar mass of a nonpolar molecular compound if 3.42 g dissolved in 41.8g benzene begins to freeze at 1.17 °c? the freezing point of pure benzene is 5.50 °c and ( k_f ), is 5.12 °c/m. a. 2.89 g/mol b. 69.2 g/mol c. 96.7 g/mol d. 126 g/mol e. 358 g/mol

Explanation:

Step1: Calculate freezing point depression

The freezing point depression \(\Delta T_f\) is the difference between the freezing point of pure solvent and the solution. So \(\Delta T_f = T_f^{\circ}-T_f = 5.50^{\circ}C - 1.17^{\circ}C=4.33^{\circ}C\).

Step2: Use freezing point depression formula to find molality

The formula for freezing point depression is \(\Delta T_f = K_f\times m\), where \(m\) is molality. Rearranging for \(m\), we get \(m=\frac{\Delta T_f}{K_f}\). Substituting values: \(m = \frac{4.33^{\circ}C}{5.12^{\circ}C/m}\approx0.8457\ m\).

Step3: Find moles of solute

Molality \(m=\frac{\text{moles of solute}}{\text{kg of solvent}}\). The mass of benzene (solvent) is \(41.8\ g = 0.0418\ kg\). Let moles of solute be \(n\). So \(n = m\times\text{kg of solvent}=0.8457\ m\times0.0418\ kg\approx0.03535\ mol\).

Step4: Calculate molar mass

Molar mass \(M=\frac{\text{mass of solute}}{\text{moles of solute}}\). Mass of solute is \(3.42\ g\), so \(M=\frac{3.42\ g}{0.03535\ mol}\approx96.7\ g/mol\).

Answer:

C. 96.7 g/mol