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
2 answers:
<span>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</span>
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.
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