Specific rotation [α] = θ/lc ------------ Eq(1)
where θ = observed rotation ,c = concentration , l = path length
Given that (R)-2-butanol is mixed with racemic-2-butanol.
Racemic mixture has equal amounts of R and S isomers, their contribution to rotation of polarized light is cancel out.
The resultant rotation is only due to (R)-2-butanol.
Hence, we have to determine the concentration of (R)-2-butanol after dilution.
Since equal volumes are mixed, volume will be doubled.
V1= V1
V1 = 2V1
C1 = 1.72 M
C2 = ?
We know that C1V1 = C2V2
1.72 M x V1 = C2 X 2V1
Then, C2 = 0.86 M
But, we need the concentration in g/mL.
Multiply the molarity with molar mass of 2-butanol to get concentration g/L .
Then,
concentration = 0.86 M x molar mass of 2-butanol
= 0.86 mol/L x 74.12 g/mol
= 63.74 g/L
= 63.74 g/ 1000 mL
= 0.06374 g/ mL
Then, determine the observed rotation using Eq (1).
Specific rotation [α] = θ/lc ------------ Eq(1)
Given that specific rotation of (R)-2-butanol, [α] = -13.9 deg ml g-1 dm-1
path length l = 1 dm
concentration c = 0.06374 g/ mL
observed rotation θ = ?
Then,
[α] = θ/lc
-13.9 deg ml g-1 dm-1 = θ/ (1 dm) x (0.06374 g/ mL)
θ = - 0.88 deg
Therefore,
observed rotation = - 0.88 deg
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