Question

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

Answer 1

Step 1: Given data

Specific rotation of sugar solution: $\alpha =0.01$

Amount of impure sugar solution: $S_{impure}=200kg/m^3$

Length of polarimeter: $l=0.25m$

Angle of rotation: $\theta =0.4rad$

Step 2: Formula used

The specific rotation of polarimeter is given by $\alpha =\frac{\theta }{lC}$, where is the concentration of the solution. Thus, $C=\frac{\theta }{l\alpha }$.

Step 3: Solution

Calculate the concentration of the solution.

$\begin{array}{rcl}C& =& \frac{\theta }{l\alpha }\\ & =& \frac{0.4}{0.25\times 0.01}\\ & =& 160kg/m^3\end{array}$

Calculate the percentage of the purity of the sugar solution.

$\begin{array}{rcl}PureSugarPercentage& =& \frac{160}{200}\times 100\\ & =& 80\%\end{array}$

Hence, the percentage of purity of sugar is the sample is $80\%$.