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The specific rotation of (S)-2- butanol is + 13.5. If 1.00 g is dissolved in 10.0 mL of ethanol and placed in a sample cell with a length of 1.00 dm, what observed rotation do you expect?

Respuesta :

Answer:

-1.35 degree

Explanation:

Given the specific rotation of (S)-2-butanol = +13.5 degree

So, the specific rotation of its enantiomer (R)-2-butanol = -13.5 degree

Now the optical rotation of a pure liquid is given by

[tex]$\theta = [\alpha]_\lambda ^t.l.d$[/tex]

Here, [tex]$\theta$[/tex]  = the optical rotation

         [tex]$[\alpha]_\lambda^t$[/tex] = specific rotation at temperature 't' and wave length 'λ'

                = -13.5 degree

         l = optical path length

           = 1 dm

          d  = density

             = 1/10

Therefore,

[tex]$\theta = -13.5 \times 1 \times \frac{1}{10}$[/tex]

    = -1.35 degree

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