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

Answer:

Specific rotation of cholesterol is -3.15°.

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

$[\alpha]_D=\dfrac{\alpha}{l.c}$

Now by putting the all values

$[\alpha]_D=\dfrac{\alpha}{l.c}$

$[\alpha]_D=\dfrac{-0.441}{1\times 0.14}$

$[\alpha]_D=-3.15$°

Specific rotation of cholesterol is -3.15°.