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?

Question

3 years ago

10

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:

Grace [21] · Answer 1 · 2022-04-22T20:38:24+03:00

Grace [21]3 years ago

5 0

<span>Î”T = i*Kb*m Where Î”T= Change in temperature and Kb= 0.512 âc/m m = Î”T/Kb = (101.56Â°C - 100.00Â°C)/(0.512Â°C/m) = 3.05 m</span>

NemiM [27] · Answer 2 · 2022-04-22T19:09:56+03:00

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.