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1 year ago

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A 6.00 g sample of an optically pure compound was dissolved in 40.0 mL of CCl4. The observed rotation was +3.30 °, measured in a

10.0 cm (1.00 dm) polarimeter tube.

1 answer:

sattari [20]1 year ago

8 0

The question is incomplete, the complete question is:

If a 6.00 g sample of an optically pure compound was dissolved in 40.0 mL of $CCl_4$ and the observed rotation was +3.30°, measured in a 10.0 cm (1.00 dm) polarimeter tube, how would one determine the specific rotation of the pure compound?

<u>Answer: </u>The specific rotation of the pure compound is $+22^o$

<u>Explanation:</u>

To calculate the specific rotation of a pure compound, we use the equation:

$[\alpha]=\frac{\alpha_{\text{observed}}}{C\times l\text{( in dm)}}$

where,

$[\alpha]$ = specific rotation of a pure compound

$\alpha_{\text{observed}}$ = observed rotation of the compound = [ex]+3.30^o[/tex]

C = concentration in g/mL = 6.00 g/40 mL = 0.15 g/mL

l = path length = 1.00 dm

Putting values in above equation, we get:

$[\alpha]=\frac{+3.30^o}{0.15\times 1.0}$

$[\alpha]=+22^o$

Hence, the specific rotation of the pure compound is $+22^o$

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