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

The specific rotation of the compound is +0.09°

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°

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