Question

(-) mandelic acid has a specific rotation of ${160}^o$. What would be specific rotation of a solution which contains $40\%$ (-) mandelic acid & $60\%$ (-) mandelic acid?

• A. $+{63}^o$
• B. $+{32}^o$
• C. $-{32}^o$
• D. $-{63}^o$

Answer 1

Answer: (C)

$-{32}^o$

Negative mandelic acid will have $-{158}^0$.The observed specific rotation will be:

$=\frac{60}{100}\times (-158)+\frac{-40}{100}\times (-158)=-{32}^0$