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A solution of 400mg of optically active 2-butanol in 10 mL of water, placed in a 20 cm cell shows an optical rotation of 40o. What is the specific rotation of this compound?

Respuesta :

ajeigbeibraheem

Answer:

Specififc rotation [∝] = 0.5° mL/g.dm

Explanation:

Given that:

mass = 400 mg

volume = 10 mL

For a solution,

The Concentration = mass/volume

Concentration = 400/10

Concentration = 40 g/mL

The path length l = 20 cm = 2 dm

Observed rotation [∝] = + 40°

Specififc rotation [∝] = ∝/l × c

where;

l = path length

c = concentration

Specififc rotation [∝] = (40 / 2 × 40)

Specififc rotation [∝] = 0.5° mL/g.dm

fichoh

Using the specific rotation formular, the specific rotation obtained for the compound ls 5°

Given the Parameters :

  • Observed rotation, [tex] \alpha = 40° [/tex]

  • Path length, l = 20 cm = 2 dm

Concentration, C

Concentration = mass × volume

Mass = 400 mg = 400 /1000 = 0.4g

Volume = 10 mL

Concentration = 0.4g × 10 mL = 4g/mL

The specific rotation formular, [tex] [\alpha] [/tex] is given by :

[tex] [\alpha] = \frac{\alpha_{observed}}{C \times l} [/tex]

[tex] [\alpha] = \frac{40°}{4 \times 2} = \frac{40°}{8} = 5° [/tex]

Therefore, the specific rotation of the compound ls 5°

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