The normal boiling point of pure benzene is found to be 80.10°C.What is the approximate molecular weight of a nonionizing substance if a solution of 3.55 g of the substance dissolved in 100.g of benzene has a normal boiling point of 80.19°C? K_b = 5.12°C/m for benzene,C₆H₆. A)20 amu B)500 amu C)2000 amu D)20,000 amu

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To find the molecular weight of the nonionizing substance, we can use the boiling point elevation formula: $\Delta T_b = i \cdot K_b \cdot m$, where $\Delta T_b$ is the change in boiling point, $i$ is the van't Hoff factor (which is 1 for a nonionizing substance), $K_b$ is the ebullioscopic constant of the solvent (5.12°C/m for benzene), and $m$ is the molality of the solution.Given:- $\Delta T_b = 80.19°C - 80.10°C = 0.09°C$- $K_b = 5.12°C/m$- Mass of solute = 3.55 g- Mass of solvent = 100 g = 0.1 kgFirst, calculate the molality ($m$) of the solution:$$0.09°C = 1 \cdot 5.12°C/m \cdot m$$$$m = \frac{0.09°C}{5.12°C/m} = 0.017578125 m$$Molality ($m$) is defined as moles of solute per kilogram of solvent, so:$$m = \frac{\text{moles of solute}}{0.1 kg}$$Rearranging to find moles of solute:$$\text{moles of solute} = 0.017578125 m \times 0.1 kg = 0.0017578125 \text{ moles}$$Finally, to find the molecular weight (MW) of the solute:$$MW = \frac{\text{mass of solute}}{\text{moles of solite}} = \frac{3.55 g}{0.0017578125 \text{ moles}} \approx 2019 \text{ amu}$$Therefore, the approximate molecular weight of the substance is 2000 amu.

The Normal Boiling Point of Pure Benzene Is Found to Be