Optical activity of an enantiomeric mixture is -12-6- and the specific rotation of - isomer is -30- The optical purity is -

Optical Purity = Total rotationSpecific rotation× 100 = 12.630× 100 = 42% ...

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Question

Optical activity of an enantiomeric mixture is $+ {12.6}^{\circ}$ and the specific rotation of $(+)$ isomer is $+ 30^{\circ}$ . The optical purity is $%$ .

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Solution

$Optical Purity = \frac{Total rotation}{Specific rotation} \times 100$
$= \frac{12.6}{30} \times 100$
$= 42 %$

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Optical activity of an enantiomeric mixture is +12.6∘ and the specific rotation of (+) isomer is +30∘. The optical purity is %.

Optical Purity=Total rotationSpecific rotation×100 =12.630×100 =42%

Optical activity of an enantiomeric mixture is $+ {12.6}^{\circ}$ and the specific rotation of $(+)$ isomer is $+ 30^{\circ}$ . The optical purity is $%$ .

$Optical Purity = \frac{Total rotation}{Specific rotation} \times 100$
$= \frac{12.6}{30} \times 100$
$= 42 %$