Answer to Question #328630 in General Chemistry for Max

Question

Calculate the boiling point of a solution made from 227 g of MgCl2 dissolved in 700. g of water. What is the boiling point of the solution? Kb = 0.512° C/m.

Answer 1

ΔT_b = iK_bC_m

i(MgCl₂) = 3

C_m = n_solute / m_solvent

C_m = n(MgCl₂) / m(H₂O) = (m(MgCl₂))/Mr(MgCl₂))) / m(H₂O) = (227 g / 94 g/mol) / 0.700 kg = 3.45 mol/kg

ΔT_b = 3 * 0.512° C/m * 3.45 mol/kg = 5.298°C

T_{b (solution) =} ΔT_b + T_{b (pure solvent)}

T_b = ΔT + T_b(H₂O) = 5.298°C + 100°C = 105.298°C