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

To find the molal concentration of glucose in the solution, we need to use the formula for boiling point elevation: ΔT = m * Kb. In this problem, the change in boiling point is 3.56°C. By rearranging the formula, we find that the molal concentration of glucose in this solution is 6.95 m.

Step-by-step explanation:

To find the molal concentration of glucose in the solution, we need to use the formula for boiling point elevation:

ΔT = m * Kb

Where ΔT is the change in boiling point, m is the molal concentration of the solute, and Kb is the molal boiling point elevation constant for water (0.51°C/m in this case).

In this problem, the change in boiling point (ΔT) is 3.56°C, which is the difference between the boiling point of the solution (103.56°C) and the boiling point of pure water (100.00°C). We can rearrange the formula to solve for m:

m = ΔT / Kb

m = 3.56°C / 0.512°C/m = 6.95 m

Therefore, the molal concentration of glucose in this solution is 6.95 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.