Question

The observed rotation of a solution of 1.2 g of a compound in 10 mL of water is 0.75 degrees. If the pathlength is 10 cm, what is the specific rotation of the compound

Answer 1

The specific rotation of the compound is +0.09°

Step-by-step explanation:

Step 1: Data given

Mass of compound = 1.2 grams

Volume water = 10 mL

pathlength = 10 cm = 1 dm

Observed rotation of the compound = 0.75 °

Step 2: Calculate the specific rotation

[ α ] = α /(c*l)

⇒ with [ α ] = the specific rotation = TO BE DETERMINED

⇒ with α = the observed rotation = 0.75 °

⇒ with c = 1.2 grams of compound / 10 mL

⇒ with l = the pathlength = 10 cm = 1 dm

[ α ] = 0.75 ° / ((1.2grams /10mL)*1 dm)

[ α ] = +0.09°

The specific rotation of the compound is +0.09°