Benzene boils at 80.10 ° C and has a molal boiling constant, k_b, of 2.53 C/m. When 2.15 g of a compound is dissolved in 20.0 g of benzene, the resulting solution has a boiling point of 81.10 ° C. What is the molar mass of the solute?

Question

Quizplus · Accepted Answer

The boiling point elevation formula is $\Delta T_b = i \cdot K_b \cdot m$, where $\Delta T_b$ is the change in boiling point, $K_b$ is the ebullioscopic constant, and $m$ is the molality of the solution. Given $\Delta T_b = 81.10 °C - 80.10 °C = 1.00 °C$, $K_b = 2.53 °C/m$, and the mass of solute (2.15 g) dissolved in 20.0 g of benzene, we can find the molar mass of the solute by rearranging the formula to solve for molality and then using the definition of molality ($molality = moles\ of\ solute / kg\ of\ solvent$). The calculation leads to a molar mass of approximately 272 g/mol.

Benzene Boils at 80