Question

What observed rotation is expected when a 1.08M solution of (R)-2-butanol is mixed with an equal volume of a .540M solution of racemic 2-butanol, and the resulting solution is analyzed in a sample container that is 1dm long? The specific rotation of (R)-2 butanol is -13.9 degrees mL/g dm.

Answer 1

[Specific rotation] = (observed rotation)/(concentration(g/mL) x length (in dm) of tube

So...
1.08 M + 1/2(0.54M) = 1.35M of R (because racemic is equal mixtures of both)

since concentration must be in g/mL we need to convert it.
MW of 2-butanol = 74.122g/mol
(1.35mol/L)(74.122g/mol)(1L/1000mL) = 0.10 g/mL

-13.9 degree mL / (g x dm)= (observed rotation) / (0.10 g/mL)(1dm)
(-13.9 degree mL / g x dm) x (0.10 g/mL)(1dm) = observed rotation
observed rotation = -1.39 degree