The observed optical rotation of a mixture of (-)-menthone and (+)-isomenthone w

Question

The observed optical rotation of a mixture of (-)-menthone and (+)-isomenthone with a concentration of 0.0662 g/ml was measured at +0.891degree in a 1 decimeter (dm) cell. Calculate the mole fraction (X) of (+ )-isomenthone using Equation 6. Plug-in the observed rotation (alpha) of + 0.891degree and the [alpha]D values in Table 1 for (-)-menthone and (+)-isomenthone. Hint, if you substitute the [alphax]D value for (+)-isomenthone for [alpha A], and the [alpha]D value for (-)-menthone for [alpha B], then the mole fraction XA calculated using Equation 6 will be the mole fraction of (+)-isomenthone. Calculate the equilibrium constant using these mole fractions.

Explanation / Answer

XA is for (+)-isomenthone.

alpha = 0.891

c = 0.0662 g/ml

l = 1 dm

alpha B = -29.6 degree

alpha A = +91.7 degree.

Substituting these values in the given equation,

alpha / (c.l) = +0.891 / (0.0662 x 1) = 13.459

XA = [13.459 -(-29.6)] / [(91.7) - (-29.6)]

= 43.059/121.3

=0.354

XA = Mole fraction of (+)-isomenthone = 0.3549

XB = Mole fraction of (-)-menthone = 1 - 0.3549 = 0.6450

Equilibrium constant = (XA) / (XB)

= 0.3459 / 0.6450 = 0.5502

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.