Question

An aqueous solution containing 6.65 g of an optically-pure compound was diluted to 2.00×102 mL with water and placed in a 10.0 cm long polarimeter tube. The measured rotation was −4.11° at 25° C. Calculate the specific rotation of the compound.

Answer 1

The specific rotation of an optically-pure compound is calculated using the observed rotation divided by the product of the path length and concentration. For this problem, the specific rotation is -123.68°·mL/g·dm.

Step-by-step explanation:

The student is asking how to calculate the specific rotation of an optically-pure compound using its observed rotation in a polarimetry experiment. To find the specific rotation [α], we use the formula:

[α] = α / (l × c)

where α is the observed rotation (-4.11°), l is the path length of the solution (10.0 cm converted to dm, hence 1.00 dm), and c is the concentration of the solution in g/mL. The solution's volume is 2.00 × 10² mL, making c = mass/volume (6.65 g / 200 mL).

Let's calculate c first:

c = 6.65 g / 200 mL = 6.65 g / 0.200 L = 33.25 g/L

Now convert this to g/mL:

c = 33.25 g/L × 1 L/1000 mL = 0.03325 g/mL

Now we can calculate the specific rotation:

[α] = -4.11° / (1.00 dm × 0.03325 g/mL)

[α] = -123.68°·mL/g·dm

The negative sign indicates that the compound is levorotatory.