Question

What observed rotation is expected when a 1.36 M solution of (R)-2-butanol is mixed with an equal volume of a 0.680 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.

please show all work

Answer 1

The solution of racemic has no contribution to the observed rotation, it only dilute the other solution.

Molar mass of 2-butanol is 74.1 g/mol

After dilution, the initial 1.36 M of the optical active compound (R)-2-butanol will be

               1.36 mol/L /2 = 0.68 mol/L =

                = 0.68 mol/L x 74.1 g/mol = 50.4 g/L = 0.0504 g/mL

The observed rotation will be

          α = α^o x c x l

              = –13.9 ^o mL g^–1 dm^–1 x 0.0504 g/mL x 1 dm =

               = - 0.700 ^o