How would you answer this question?

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anondukie

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“A scientist takes a 0.5 M solution of an unknown pure dextrorotatory organic molecule and places it in a test tube with a diameter of 1 cm. He observes that a plane of polarized light is rotated 12° under these conditions. What is the specific rotation of this molecule?”

Excerpt From: Kaplan. “Kaplan MCAT Organic Chemistry Review: Created for MCAT 2015 (Kaplan Test Prep).”

Can you convert from M to g/mL without being giving additional information?

Thanks!

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“A scientist takes a 0.5 M solution of an unknown pure dextrorotatory organic molecule and places it in a test tube with a diameter of 1 cm. He observes that a plane of polarized light is rotated 12° under these conditions. What is the specific rotation of this molecule?”

Excerpt From: Kaplan. “Kaplan MCAT Organic Chemistry Review: Created for MCAT 2015 (Kaplan Test Prep).”

Can you convert from M to g/mL without being giving additional information?

Thanks!
Normally in order to determine specific rotation you conduct an experiment and use the equation:

[α]LT = α/lc

Where [α]LT = specific rotation in degrees. (The correct units are deg cm2 g-1, but are usually just given as degrees). L is the wavelength of light used for the observation (usually 589 nm, the D line of a sodium lamp unless otherwise specified. T is the temperature in Celsius.

This value is characteristic for a given compound, just like the melting point. Concentration and solvent data is included if relevant.

Example: (c = 10, CH3OH) after the rotation means the specific rotation was determined at a concentration of 10 g in 100 ml-1 in methanol.

  • α = observed rotation in degrees.
  • l = cell path length in decimeters. (1 decimeter = 1 dm = 10 cm. A standard polarimeter tube is 1o cm in length.)
  • c = concentration in g/ml for a pure liquid compound, or g/100 ml for a solution.

As far as what's required on the MCAT, I'm not sure how you can get the information you need with what they have provided. Does the book offer an explanation or solution?
 
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The specific rotation equation is as follows:

[α]= α observed/ c x l

With α observed as the given 12 degrees
c as concentration and l as test tube length

From here on out, this is simple plug and chug. Convert 1cm to dm which is 0.1. Mutiply 0.5 x 0.1 to get 0.05. Next, Divide 12 by 0.05 to get 240 degrees, and that is your answer.
 
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The specific rotation equation is as follows:

[α]= α observed/ c x l

With α observed as the given 12 degrees
c as concentration and l as test tube length

From here on out, this is simple plug and chug. Convert 1cm to dm which is 0.1. Mutiply 0.5 x 0.1 to get 0.05. Next, Divide 12 by 0.05 to get 240 degrees, and that is your answer.

Conceptually correct but not entirely applicable to this problem. Concentration must be normalized to a reference state, which is in g/mL. Since this problem gives concentration in terms of molarity without specifying molar weight, it's not possible to get to g/mL.
 
The explanation in the book just uses 0.5 M as the concentration without converting to g/mL. It must be an oversight. All of you provided interesting responses. Thank you!
 
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