Pure (R )-mandelic acid has specific rotation of -154∘ and sample contains 60% of (R) isomer and 40% of its enantiomers. Specific rotation [α] of mixture is

: 60%-40%=20% exaticmatric across 20=[α]mtx−154×100=−3160%-40%=20%

Pure (R )-mandelic acid has specific rotation of -154∘ and sample contains 60% of (R) isomer and 40% of its enantiomers. Specific rotation [α] of mixture is

: 60%-40%=20% exaticmatric across 20=[α]mtx−154×100=−3160%-40%=20%

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Pure (R )-mandelic acid has specific rotation of $- 154^{\circ}$ and sample contains 60% of (R) isomer and 40% of its enantiomers. Specific rotation $[α]$ of mixture is

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$- 62^{\circ}$

$+ 31^{\circ}$

$- 31^{\circ}$

$+ 62^{\circ}$

answer is B.

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Detailed Solution

: $60 % - 40 % = 20 %$ exaticmatric across $20 = \frac{{[α]}_{m t x}}{- 154} \times 100 = - 31$
$60 % - 40 % = 20 %$

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Pure (R )-mandelic acid has specific rotation of -154∘ and sample contains 60% of (R) isomer and 40% of its enantiomers. Specific rotation [α] of mixture is

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Pure (R )-mandelic acid has specific rotation of $- 154^{\circ}$ and sample contains 60% of (R) isomer and 40% of its enantiomers. Specific rotation $[α]$ of mixture is