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.

Question

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.

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Khoii Minh 6 months 2021-07-17T23:41:42+00:00 1 Answers 2 views 0

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    2021-07-17T23:42:46+00:00

    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?

    Answer: The specific rotation of the pure compound is +22^o

    Explanation:

    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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