Question

The observed rotation of 2.0g of a compound in a 50ml of solution in a polarimeter tube 20cm long is +138°, what is the specific rotation of the compound?

Answer 1

Specific rotation is equal to: $[\alpha]=\frac{\alpha}{l\cdot c}$

where $\alpha$ is the measured rotation in degrees, l is the path length in decimeters, c is the concentration in g/mL. Thus,

$[\alpha]=\frac{\alpha}{l\cdot c}=\frac{\alpha \cdot V}{l\cdot m}=\frac{+138^{\circ} \cdot 50 mL}{2dm \cdot 2.0 g}= +1725^{\circ}$