Question

What observed rotation is expected when a 1.46 M solution of (R)-2-butanol is mixed with an equal volume of a 0.730 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 – In this one there is given like 1.46 M solution of (R)-2-butanol is mixed with an equal volume of a 0.730 M solution of racemic 2-butanol, means

concentration C = 1.46 M +0.730 / 2

                          = 1.825 M

l = 1 dm, [α] =   -13.9 degrees mL g^-1 dm^-1

We know formula

Observed rotation, α = [α] * C* l

mass of (R)-2-butanol = 1.825 mole * 74.122 g/mol

                                     = 135.27 g

So in the g/mL = 135.27 g / 1000 mL

                         = 0.1353 g/mL

So, observed rotation ,α = [α]_λ^T * l *C

                                        = -13.9^o mL/g.dm * 1 dm * 0.1353 g/mL

                                       = -1.88^o

So, -1.88^o observed rotation is expected when a 1.46 M solution of (R)-2-butanol is mixed with an equal volume of a 0.730 M solution of racemic 2-butanol