Question

Specific rotation of sugar solution is $0.5 \, deg \, m^{2} \, kg^{-1}. \,200 kgm^{-3}$ 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. 0.2
• B. 80%
• C. 0.95
• D. 0.89

Answer 1

The Correct Option is 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 $ $\therefore=\frac{19}{0.20 \times 0.5}=190 kg - m ^{-3}$ The sugar sample dissolved in $a \,m ^{3}$ of water is $200\, kg$ in which $190\, kg$ is pure sugar. Therefore, purity is $\frac{190}{200} \times 100=95 \%$