The (-) enantiomer of a compound has a specific rotation of [a] = 123o. What rotation, a, will be observed for a sample of this pure enantiomer at a concentration of 0.25 g/10 mL, when measured in the polarimeter in a two dm sample tube?



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[a] = a/(c x l)

 


The specific rotation, [a] = 123o

The concentration of 0.25 g/10 mL is equivalent to 0.025 g/mL;   c = 0.025 g/mL

The length of the analysis tube is two dm;  l = 2.0 dm





















































Solution:

The specific rotation, [a] = 123o

The concentration of 0.25 g/10 mL is equivalent to 0.025 g/mL;   c = 0.025 g/mL

The length of the analysis tube is two dm;  l = 2.0 dm



The specific rotation is therefore:


[a] = 123o =  a /(0.025 g/mL  x 2.0 dm)


a = 6.15o g-1 mL-1 dm-1


 

























































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