Question

When 0.14 g of cholesterol is dissolved in 1.00 mL of ether and placed in a sample cell 10.0 cm in length, the observed rotation at 20°C (using the D line of sodium) is -0.441°. Calculate the specific rotation of cholesterol.

Answer 1

Specific rotation of cholesterol is -3.15°.

Step-by-step explanation:

Given that

Path length = 10 cm = 1 dm.

Observed rotation ,α= - 0.441 °

m= 0.14 g

V= 1 ml

The concentration of cholesterol ,c= m/V g/ml

c = 0.14 / 1 = 0.14 g/ml

We know that specific rotation given as

`[\alpha]_D=(\alpha)/(l.c)`

Now by putting the all values

`[\alpha]_D=(\alpha)/(l.c)`

`[\alpha]_D=(-0.441)/(1* 0.14)`

`[\alpha]_D=-3.15`°

Specific rotation of cholesterol is -3.15°.