A polarimetric study of sucrose inversion gives rotations +24.0 deg (t = 0), +12.0 deg (t = 15 min) and -8.0 deg (infinite time). Assuming first-order kinetics and ln10=2.303, find the time in minutes (to one decimal place) for 90% of the sucrose to be inverted.

JEE Advanced levelNumerical answerDifficulty 3 of 3ck150

A polarimetric study of sucrose inversion gives rotations $+ 24. 0^{\circ}$ (t = 0), $+ 12. 0^{\circ}$ (t = 15 min) and $- 8. 0^{\circ}$ (infinite time). Assuming first-order kinetics and $ln 10 = 2.303$ , find the time in minutes (to one decimal place) for 90% of the sucrose to be inverted.

Numerical answer type, answer in min, tolerance 0.4.

k = \frac{1}{15} ln \frac{32}{20} = 0.03133 mi n^{- 1}

. For 90% inversion, 10% of

(r_{0} - r_{\infty})

remains, so

t = \frac{2.303}{k} = 73.5

min.