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A 0.1 m solution of an enantiomerically pure chiral compound (A) has an observed rotation of 0.20 in a 1dm sample tube. The molecular mass of the compound is 150. What is the specific rotation of D? The concentration of the solution is 0.05g/ml.Can’t get the specific angle of rotation.

A 0.1 m solution of an enantiomerically pure chiral compound (A) has an observed rotation of 0.20 in a 1dm sample tube. The molecular mass of the compound is 150.
What is the specific rotation of D?
The concentration of the solution is 0.05g/ml.Can’t get the specific angle of rotation.

Grade:12th pass

1 Answers

Neeti
571 Points
8 years ago
specific rotation of a pure chiral compound is given by : \alpha / l x c 
where \alpha is observed rotation in degrees.
l is cell path length in decimeters. 
and c is concentration in g/ml
 
So according to the formula it should be : 0.20 / 1 x 0.05  ( assuming that 0.20 is the observed totation in degrees)  =  4 
 
Feel free to ask any follow up question and approve the answer if you like it :)

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