What is the boiling point of a solution of 11.0 g of lactose (C₁₂H₂₂O₁₁) in 145.0g of water (K_b (H₂O) = 0.52 ^$\circ$C/m) A) 0.1^$\circ$C B) 99.9 ^$\circ$C C) 100.1^$\circ$C D) 102.2 ^$\circ$C E) 105.7 ^$\circ$C

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The boiling point elevation can be calculated using the formula ΔTb = iKbm, where ΔTb is the boiling point elevation, i is the van't Hoff factor (which is 1 for non-electrolytes like lactose), Kb is the ebullioscopic constant of the solvent (water in this case, 0.52 °C/m), and m is the molality of the solution. First, calculate the molar mass of lactose (C12H22O11) which is 342.30 g/mol. Then, find the molality: molality = moles of solute/kg of solvent = (11.0 g / 342.30 g/mol) / (145.0 g / 1000 g/kg) = 0.209 m. Finally, calculate the boiling point elevation: ΔTb = 1 * 0.52 °C/m * 0.209 m = 0.10868 °C. The boiling point of the solution is then 100 °C (boiling point of pure water) + 0.10868 °C ≈ 100.1 °C.

What Is the Boiling Point of a Solution of 11 ∘\circ∘

What Is the Boiling Point of a Solution of 11 $\circ$