Question

Problem 4.5 ine concentration of cholesterol dissolved in chloroform is 6.15 g
per 100 ml of solution. (a) A portion of this solution in a 5-cm polarimeter tube causes
an observed rotation of -1.2, Calculate the specific rotation of cholesterol. (b) Pre-
dict the observed rotation iſ the same solution were placed in a 10-cm tube (c) Pre-
dict the observed rotation of 10 ml of the solution were diluted to 20 ml and placed in 5 cm tube​

Answer 1

Final answer:

(a) Specific rotation of cholesterol [α] = - 36.92°      

(b)Observed rotation of cholesterol $\alpha_{obs}= -2.399$

(c) Observed rotation of cholesterol $\alpha_{obs}=-0.599$

Given that: We are given concentration of cholesterol dissolved in chloroform is 6.15 g per 100 ml of solution.

To find: We have to find,

(a) Specific rotation of cholesterol, when solution in a 5 cm polarimeter tube causes an observed rotation of - 1.2°.

(b) Observed rotation iſ the same solution were placed in a 10 cm tube.

(c) Observed rotation of 10 ml of the solution were diluted to 20 ml and placed in 5 cm tube​.

Explanation:

• Specific rotation of an optically active compound, at a specific temperature and for specific wavelength of light, is defined as the observed rotation in degrees when plane-polarised light is passed through one decimeter (or 10 cms) of the solution of the substance having a concentration of one $\frac{gram}{mL}$.

I.e. Specific rotation $[\alpha] = \frac{\alpha _{obs}}{c*l}$

where $\alpha _{obs}$ = Observed rotation in degrees

l = Length of the solution in dm.

c = Concentration in $\frac{g}{mL}$

• (a)  

$\alpha _{obs}$  = - 1.2°

c =   $\frac{6.15 g}{100 mL}$

l = 5 cm = 0.5 dm

Specific rotation of cholesterol,

$[\alpha] = \frac{\alpha _{obs}}{c*l} = \frac{ -1.2}{(\frac{6.5}{100})*0.5 }$

[α] = - 36.92°      

• (b)

[α] = - 36.92°  

c = $\frac{6.15 g}{100 mL}$

l = 10 cm = 1 dm

Substitute these values in specific rotation equation.

$- 36.92 =\frac{ \alpha_{obs}}{(\frac{6.5}{100})*1}$

$- 36.92 =\frac{ \alpha_{obs}}{(\frac{6.5}{100})}$

$\alpha_{obs}= - 36.92 *(\frac{6.5}{100}) = - 36.92 * 0.065 = -2.399\\\\\alpha_{obs} =- 2.399$

Observed rotation of cholesterol $\alpha_{obs}= -2.399$

• (c)

[α] = - 36.92°      

When 10 ml of the solution were diluted to 20 ml, c = $\frac{6.15 g}{200 mL}$

l = 5-cm = 0.5 dm

Substitute these values in specific rotation equation.

$- 36.92 =\frac{ \alpha_{obs}}{(\frac{6.5}{200})*0.5}$

$- 36.92 =\frac{ \alpha_{obs}}{0.01625}$

$\alpha_{obs}= - 36.92 * (0.01625)\\\\\alpha_{obs}=-0.599$

Observed rotation of cholesterol $\alpha_{obs}=-0.599$

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