Answer to Question #340280 in General Chemistry for ROD.B

Question

78.0 g of nonionizing C₆H₁₂O₆ are dissolved in 300 g of H₂O.

What is the change in boiling temperature in ⁰C, (ΔT_b)?

Answer 1

∆T = imK_b

∆T = change in temperature

i = van't Hoff factor = 1 for glucose since it does not ionize or dissociate. It is a single particle.

K_b = 0.512 ⁰C/m

m = molality = moles solut/kg solvent = (78.0 g/180 g/mol) / 0.3 kg = 1.44 m

∆T = 1 * 1.44 m * 0.512 ⁰C/m = 0.74 ⁰C