If 14.2 g of Al(NO₃)₃ is dissolved in 655g of water, what is the boiling point of the solution (assume 100% dissociation, K_b (H₂O) = 0.52 ^$\circ$C/m) A) 99.8^$\circ$C B) 99.9^$\circ$C C) 100.1^$\circ$C D) 100.2^$\circ$C E) 100.4^$\circ$C

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Quizplus · Accepted Answer

The boiling point elevation can be calculated using the formula ΔT_b = i * K_b * m, where i is the van't Hoff factor (4 for Al(NO3)3), K_b is the ebullioscopic constant (0.52 °C/m), and m is the molality. First, calculate the molality: moles of Al(NO3)3 = 14.2 g / 213 g/mol = 0.0667 mol; molality = 0.0667 mol / 0.655 kg = 0.1018 m. ΔT_b = 4 * 0.52 °C/m * 0.1018 m = 0.211 °C. The boiling point of the solution is 100 °C + 0.211 °C = 100.2 °C.

If 142 G of Al(NO3)3 Is Dissolved in 655g of Water

If 142 G of Al(NO₃)₃ Is Dissolved in 655g of Water