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The optical rotation of a solution of pure natural camphor is found to be $+ {5.76}^{0}$ Under the following conditions the concentration $= 0.13 g / m l$ , length of polarimeter $= 1 d m,$ Wavelength = sodium D line, $T = 25^{0} C$ . The specific rotation of camphor is

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$- {44.3}^{0}$

$+ {44.3}^{0}$

$- {26.7}^{0}$

$+ {26.7}^{0}$

answer is A.

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Detailed Solution

$α$ specific rotation of campour
$α = \frac{o b s e r v e d r o t a t i o n}{C \times l}$
$\begin{array}{l} = \frac{5.76}{0.13 \times 1} \\ = + 44.3 \end{array}$

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