QQuestionChemistry
QuestionChemistry
(+)-Tartaric acid has a specific rotation of + 12.0°. What is the specific rotation of a mixture of 75% (+)-tartaric acid and 25% (-)-tartaric acid?
A. + 4.0°
B. + 6.0°
C. + 8.0°
D. + 9.0°
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Answer
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Step 1: Understand the problem
In this problem, $\alpha_{1} = +12.0°$, $m_{1} = 0.75 \times m_{total}$, and $\alpha_{2} = -12.0°$, $m_{2} = 0.25 \times m_{total}$.
We are given the specific rotation of pure (+)-tartaric acid and the percentage of a mixture containing (+)-tartaric acid and (-)-tartaric acid. We need to find the specific rotation of the mixture. The specific rotation of a chiral molecule is a measure of its optical activity, which describes the ability of the molecule to rotate plane-polarized light. When two enantiomers are present in a mixture, their optical activities partially cancel each other out. The specific rotation of the mixture can be calculated using the following formula: Where:
Step 2: Calculate the specific rotation of the mixture
[\alpha]_{mixture} = +6.0°
Plug the given values into the formula:
Final Answer
The specific rotation of the mixture is + 6.0°.
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