What is the boiling point of a solution of 1.00 g of naphthalene dissolved in 94.0 g of toluene? The — Solutions and Colligative Properties Chemistry Question

💡 Solution & Explanation

$K_{x,b}$ relates boiling point elevation to mole fraction: $\Delta T_b = K_{x,b} \times X_{solute}$. Naphthalene ($C_{10}H_8$) mass $= 1.00$ g, $M = 128 \text{ g/mol}$, so $n_{solute} = 1.00/128 = 0.00781$ mol. Toluene ($C_7H_8$) mass $= 94.0$ g, $M = 92 \text{ g/mol}$, so $n_{solvent} = 94.0/92 = 1.0217$ mol. $X_{solute} = 0.00781 / (0.00781 + 1.0217) = 0.007586$. $\Delta T_b = 32.0 \times 0.007586 = 0.242^\circ C$. New BP $= 110.75 + 0.242 = 110.992^\circ C \approx 111.0^\circ C$. Therefore, correct answer is B.