Q. Specific rotation of sugar solution is 0.5 deg $m^{2}k/g.\ \ 200\ kgm^{- 3}$ of impure sugar solution is taken in a sample polarimeter tube of length 20 cm and optical rotation is found to be $19{^\circ}.$ The percentage of purity of sugar is
A) 20%
B) 80%
C) 95%
D) 89%
Explanation
(c) The strength of solution is given by
$c = \frac{\theta}{l \times s}$
Where the symbols have their usual meanings.
Here, $\theta = 19{^\circ},\ l = 20\ cm = 0.20\ m$
$S = 0.5\deg{m^{2}{kg}^{- 1}}$
$\therefore c = \frac{19}{0.20 \times 0.5} = 190\ kg - m^{- 3}$
The sugar sample dissolved in a${\ m}^{3}\ $of water is 200 kg in which 190 kg is pure sugar.
Therefore, purity is $\frac{190}{200} \times 100 = 95\%$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.