Question

What observed rotation is expected when a 1.72 M solution of (R)-2-butanol is mixed with an equal volume of a 0.860 M solution of racemic 2-butanol, and the resulting solution is analyzed in a sample container that is 1 dm long? The specific rotation of (R)-2-butanol is –13.9 degrees mL g–1 dm–1.

Answer 1

Answer:

α = -0,885°

Explanation:

The specific rotation [α] is defined as:

[α] = α/ c×l (1)

Where α is the observed rotation, c is concentration in g/mL, and l is optical path length in dm.

A racemic mixture has no optical activity. As the solution of (R)-2-butanol is mixed with an equal volume of the racemic mixture the concentration of the original solution decreases twice. That is 1,72M/2 = 0,86M. This concentration in g/mL is:

0,86mol/L×$\frac{1L}{1000mL}$×$\frac{74,122g}{1mol}$= 0,0637 g/mL

Replacing in (1):

$-13,9mL.g^{-1}.dm^{-1}=\frac{a}{0,0637g/mL*1dm}$

α = -0,885°

I hope it helps!