Specific rotation of sugar solution is `0.01 SI` units. `200 kg-m^(-3)` of impure sugar solution is taken in a polarimeter tube of length `0.25m` and an optical rotation of `0.4` rad is observed. The percentage of purity of sugar in the sample is

Options

A. 0.11
B. 0.2
C. 0.8
D. 0.89

Answer

C

Explanation

Specific rotation,
`alpha=("angle of rotation" (radians),theta)/("length of the tube xx concentration,C")`  
`therefore`   `C=(theta)/(alpha.l)`
Here I = 0.25 m, = 0.01, 0 = 0.4 rad,
`rArr`   `C=(0.4)/(0.01xx0.25)=160kg//m^(3)` But 200 kg/ml was taken
`therefore`  purity =`(160)/(200)xx100=80%`