Question

A solution of water (kb=0.512 ∘c/m) and glucose boils at 103.56 ∘c. what is the molal concentration of glucose in this solution? assume that the boiling point of pure water is 100.00 ∘c. express your answer to three significant figures and include the appropriate u

Answer 1

ΔT = i*Kb*m Where ΔT= Change in temperature and Kb= 0.512 âc/m m = ΔT/Kb = (101.56°C - 100.00°C)/(0.512°C/m) = 3.05 m

Answer 2

Answer is: molal concentration of glucose in this solution is 6.953 mol/kg.

Kb(H₂O) = 0.512 °C/m.

T(glucose) = 103.56 °C.

T(H₂O) = 100°C.

ΔT = 103.56°C - 100°C.

ΔT = 3.56°C.

ΔT = m(glucose) · Kb(H₂O).

m(glucose) = ΔT ÷ Kb(H₂O).

m(glusose) = 3.56°C ÷ 0.512 °C/m.

m(glucose) = 6.953 mol/kg.

m - molal concentration.