Asolution of water (kb=0.512 ∘c/m) and glucose boils at 101.56 ∘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 units.

Answer is: molal concentration of glucose in this solution is 3,04 mol/kg (molal).
Kb(H₂O) = 0,512 °C/m.
T(glucose) = 101,56 °C.
T(H₂O) = 100°C.
ΔT = 101,56°C - 100°C.
ΔT = 1,56°C.
ΔT = m(glucose) · Kb(H₂O).
m(glucose) = ΔT ÷ Kb(H₂O).
m(glusose) = 1,56°C ÷ 0,512 °C/m.
m(glucose) = 3,04 mol/kg.
m - molal concentration.

Idon't see the pic?